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Where the coefficients come from

◆FRAME

Multiply $(a+b)^3$ out the long way, without collecting terms yet.

$(a+b)(a+b)(a+b)$ is three brackets. To build one term you take a letter from each bracket and multiply them together. Every choice of letters gives a term, so there are $2 \times 2 \times 2 = 8$ terms before any collecting.

List the ones that give $a^2 b$ — two a factors and one b. The b can come from the first bracket, or the second, or the third. Three terms, so the collected coefficient is 3.

Now $a b^2$, two b factors and one a. The single a can come from any of the three brackets. Three terms again, coefficient 3.

And $a^3$? Every bracket must hand over its a. One way only, so the coefficient is 1. The same goes for $b^3$.

Collecting gives

$(a+b)^3 = a^3 + 3a^2 b + 3a b^2 + b^3,$

and every coefficient turned out to be a COUNT — the number of ways of choosing which brackets contributed the b.

Counting which $r$ things to choose out of $n$ is exactly $C(n,r)$, from the last chapter. Here $C(3,1) = 3$, and it matches.

The coefficients are not a separate fact to memorise. They are combination counts, and Pascal’s Triangle is a fast way of reading them rather than a second rule.

All eight products of (a+b)(a+b)(a+b) written out and sorted by how many b’s they carry: the size of each column is that term’s coefficient.
The 16 ways (a+b) to the fourth can hand over an a or a b, sorted by how many b’s: the column heights are the coefficients.
The two exponents of (a+b) cubed on facing rails, running opposite ways, with their sum on a third rail underneath.

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The binomial theorem

▲CONCEPT

The pattern from the opener, stated in general. For any positive integer n,

$(a+b)^n = \sum_{r=0}^n C(n,r) a^{n-r} b^r \cdot$

Take it apart. The sum runs over $r = 0, 1, 2, \ldots, n$, so the expansion has exactly $n+1$ terms — one more than the power, which is worth keeping as a check.

Within the term for a given r, the exponent of b is r and the exponent of a is $n-r$. Those two always add to n, in every term. That is the second useful check.

And the coefficient is $C(n,r)$: the number of ways to choose which r of the n brackets hand over their b.

Read the extremes to see that it fits. At $r = 0$ the term is $C(n,0) a^n = a^n$, which is a taken from every bracket. At $r = n$ it is $C(n,n) b^n = b^n$. Both coefficients are 1, as they must be — there is only one way to take everything from the same letter.

*Two checks before any arithmetic: $n+1$ terms, and exponents summing to n in each of them.*

The seven terms of (a+b) to the sixth, one column per r: the coefficients read 1, 6, 15, 20, 15, 6, 1 in either direction, and only the middle one is unpaired.
Worked example

Expand $(x+y)^4$

  1. $n = 4$, $a = x$, $b = y$
    Expect $4 + 1 = 5$ terms.
  2. $C(4,0)x^4 + C(4,1)x^3 y + C(4,2)x^2 y^2 + C(4,3)x y^3 + C(4,4)y^4$
    Write the skeleton first, letting r run from 0 to 4 and keeping the exponents adding to 4 — doing both the powers and the coefficients at once is where terms get dropped.
  3. $C(4,0) = 1$, $C(4,1) = 4$, $C(4,2) = 6$, $C(4,3) = 4$, $C(4,4) = 1$
    Fill in the coefficients.
  4. $(x+y)^4 = x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4$
    Combine the skeleton and the coefficients.
  5. $4+0$, $3+1$, $2+2$, $1+3$, $0+4$
    Run the checks: five terms, matching $n+1$, and in each term the exponents add to 4.
  6. $1, 4, 6, 4, 1$
    The coefficients read the same forwards and backwards — the symmetry $C(n,r) = C(n, n-r)$ from the last chapter, turning up in every expansion.
The five terms of (x+y) to the fourth, one column each: term number, r, the coefficient, and the powers it multiplies.

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The same counts, read off a triangle

●KEY-TERM

Every coefficient in this chapter is some $C(n,r)$, and there is a way to get a whole row of them without evaluating any.

PASCAL’S TRIANGLE arranges the values in a grid. Row n holds the $n+1$ coefficients of $(a+b)^n$, left to right, from $C(n,0)$ through to $C(n,n)$.

The first rows are 1; then 1, 1; then 1, 2, 1; then 1, 3, 3, 1; then 1, 4, 6, 4, 1. That last one is the row which just produced the expansion of $(x+y)^4$.

The construction rule is the point of the thing. Every entry is the SUM of the two entries diagonally above it, and every row starts and ends with a 1.

That rule is not a new fact either. It is Pascal’s rule from the last chapter, $C(n,r) + C(n, r-1) = C(n+1, r)$, drawn as a picture. The two entries above a given place are exactly the two combination counts on the left of that identity.

Small n and a whole row wanted: build the triangle. Large n, or one term out of many: use the formula.

Pascal’s Triangle, rows 0 to 6, with row 5’s 10 marked as the 4 and 6 above it, added.
Worked example

Build row 5 of Pascal’s Triangle

  1. Row 4 is $1, 4, 6, 4, 1$
    Start from the row already known.
  2. $1+4 = 5$, then $4+6 = 10$, then $6+4 = 10$, then $4+1 = 5$
    Each entry after the first sums the two above it — build each row from the one above, never from scratch.
  3. Row 5 is $1, 5, 10, 10, 5, 1$
    Count the entries: six, which is $n+1$ with $n = 5$. Correct.
  4. $(a+b)^5 = a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5a b^4 + b^5$
    These are the coefficients of $(a+b)^5$.
  5. $C(5,2) = (5 \times 4)/(2 \times 1) = 10$
    Check one entry against the formula — it agrees.
  6. $C(5,2)$ and $C(5,3)$
    The row is symmetric again, and it has to be: the two count the same thing from opposite ends.
Row 5 built from row 4: each interior entry is the pair above it added, and the two ends have no pair, so they stay 1.

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One term at a time

▲CONCEPT

Expanding everything to reach one term is wasteful, and for a large power it is not practical at all. There is a formula for a single term.

The $(r+1)$th term of $(a+b)^n$, written $T_{r+1}$, is $T_{r+1} = C(n,r) a^{n-r} b^r$.

That is one summand of the theorem, lifted out and given a name.

The indexing needs care, and it is the source of the commonest error in this chapter. The formula produces the $(r+1)$th term, not the rth. So $r = 0$ gives $T_1$, the FIRST term. $r = 1$ gives $T_2$. And so on up.

The working rule: to find the kth term, set $r + 1 = k$, so $r = k - 1$. Subtract one before substituting, every time.

The reason for the offset is that r counts how many b factors a term carries, and the first term carries none. Counting from zero is natural for r and unnatural for term positions, and the formula sits across both conventions.

*The formula is indexed by r and the question is indexed by position. They differ by one, so substitute $r = k-1$.*

The six terms of (a+b) to the fifth over two rails: term number 1 to 6 above, r 0 to 5 below.
Worked example

Find the 3rd term of $(x+2)^6$

  1. $T_3 = T_{r+1}$, so $r + 1 = 3$ and $r = 2$
    Position first — subtracting that one is the whole of the care needed here.
  2. $n = 6$, $a = x$, $b = 2$
    Identify the pieces.
  3. $T_3 = C(6,2) x^{6-2} 2^2 = C(6,2) x^4 2^2$
    Substitute into $T_{r+1} = C(n,r) a^{n-r} b^r$.
  4. $C(6,2) = (6 \times 5)/(2 \times 1) = 15$, $2^2 = 4$
    Evaluate.
  5. $T_3 = 15 \times 4 \times x^4 = 60 x^4$
    Multiply through.
  6. $n - r = 6 - 2 = 4$
    Sanity-check the exponent of x. A slip putting $r = 3$ would silently give the 4th term instead, which is why the check is worth the second it takes.
One column read downwards is the whole method: term number 3 fixes r at 2, and r = 2 selects the coefficient 15.
Worked example

Find the term independent of x in $(x + 1/x)^6$

  1. $T_{r+1} = C(6,r) x^{6-r} (1/x)^r$
    Write the general term with $n = 6$, $a = x$, $b = 1/x$ — independent of x means the power of x must come out as zero, so the real question is which r makes that happen.
  2. $(1/x)^r = x^{-r}$, so $x^{6-r} x^{-r} = x^{6-2r}$
    Simplify the powers of x.
  3. $T_{r+1} = C(6,r) x^{6-2r}$
    Rewrite the general term with the combined exponent.
  4. $6 - 2r = 0$, so $r = 3$
    Set that exponent to zero and solve, rather than hunting term by term.
  5. $C(6,3) x^0 = 20 \times 1 = 20$
    Substitute back — the constant term is 20. r came out a whole number between 0 and 6, which it has to; a fraction or an out-of-range value would mean no such term exists, and that is a real answer, not a failed attempt.
The power of x falls by 2 at every step, so the term independent of x is found by solving 6 - 2r = 0, not by trying terms one at a time.

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The middle term

▲CONCEPT

The expansion of $(a+b)^n$ has $n+1$ terms, and how many sit in the middle depends on whether that count is odd or even.

If n is EVEN then $n+1$ is odd, and an odd list has exactly ONE middle entry. It is the $(n/2 + 1)$th term.

If n is ODD then $n+1$ is even, and an even list has no single middle. It has TWO: the $((n+1)/2)$th and $((n+3)/2)$th terms.

Rather than memorise those four expressions, rebuild them. With $n = 6$ there are 7 terms, so the middle one is the 4th, and $6/2 + 1 = 4$ agrees. With $n = 7$ there are 8 terms, so the middle two are the 4th and 5th, and $(7+1)/2 = 4$ and $(7+3)/2 = 5$ agree.

Once the POSITION is known, nothing new is needed. Convert it to r by subtracting one, then feed it to the general term formula.

Count the terms, find the middle position, subtract one to get r. The formulas are that count written out.

Two term rails: seven chips for n = 6 with one middle chip marked, eight chips for n = 7 with two.
Worked example

Find the middle term of $(x+2)^6$

  1. $n = 6$ (even), so $6 + 1 = 7$ terms
    Even n gives an odd number of terms — exactly one middle one.
  2. $n/2 + 1 = 6/2 + 1 = 4$
    Its position is half of n, plus one. The 4th term is wanted.
  3. $r + 1 = 4$, so $r = 3$
    Convert position to r.
  4. $T_4 = C(6,3) x^3 2^3$
    Substitute into $T_{r+1} = C(n,r) a^{n-r} b^r$ with $a = x$ and $b = 2$.
  5. $C(6,3) = 20$, $2^3 = 8$, so $T_4 = 20 \times 8 \times x^3 = 160 x^3$
    Evaluate.
  6. $n - r = 3$
    Two checks: the exponent of x matches, and with 7 terms the 4th genuinely sits three in from each end.
The next worked example flips the parity: an odd exponent gives an even number of terms, so the split can no longer land on one term, and the expansion carries two middle terms instead of a single one. It is the term count that decides which case applies, not anything about the coefficients themselves.
Worked example

Find the middle terms of $(x+2)^7$

  1. $n = 7$ (odd), so $7 + 1 = 8$ terms
    Odd n gives an even number of terms and no single middle — two terms share the centre, and each is worked separately.
  2. $(7+1)/2 = 4$ and $(7+3)/2 = 5$
    Their positions are the 4th and 5th terms.
  3. $r + 1 = 4$, so $r = 3$; $T_4 = C(7,3) x^{7-3} 2^3 = C(7,3) x^4 2^3$
    Take the 4th term.
  4. $C(7,3) = 35$, $2^3 = 8$, so $T_4 = 35 \times 8 \times x^4 = 280 x^4$
    Evaluate it.
  5. $r + 1 = 5$, so $r = 4$; $T_5 = C(7,4) x^3 2^4$
    Take the 5th term.
  6. $C(7,4) = 35$, $2^4 = 16$, so $T_5 = 35 \times 16 \times x^3 = 560 x^3$
    Evaluate it.
  7. $C(7,3) = C(7,4) = 35$
    The symmetry again — a useful sign the two positions were picked correctly, though only the coefficients coincide.
Eight terms leaves no single middle: three on each side and two in the centre. The two coefficients are both 35; the two answers are not the same.

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When the sign is a minus

▲CONCEPT

Nothing new is needed for a minus sign. Substitute.

Write $a - b$ as $a + (-b)$ and put it through the theorem:

$(a-b)^n = \sum_{r=0}^n C(n,r) a^{n-r} (-b)^r \cdot$

The only piece that changed is $(-b)^r$, and its behaviour is simple. A negative number raised to an EVEN power is positive; raised to an ODD power it is negative.

So terms with even r keep their sign and terms with odd r flip. Since r runs $0, 1, 2, 3, \ldots$, the signs run plus, minus, plus, minus and so on, alternating from the first term.

The coefficients themselves do NOT change. $C(n,r)$ counts choices and knows nothing about signs, so row 4 of Pascal’s Triangle is still $1, 4, 6, 4, 1$ for $(a-b)^4$. It is only the signs in front that alternate.

That also settles the last term. It is the one with $r = n$, so an $(a-b)^n$ expansion ends on a plus when n is even and a minus when n is odd.

Same coefficients, alternating signs, starting with a plus. Both facts come out of the one substitution.

Four rails: r, the coefficients of (a+b) to the fifth, the sign of each term, and the coefficients of (a-b) to the fifth.
Worked example

Expand $(x-2)^4$

  1. $a = x$, $b = 2$
    The minus is handled by alternating signs, not by changing a or b.
  2. Row 4: $1, 4, 6, 4, 1$
    Pascal’s Triangle gives the coefficients — unchanged by the minus.
  3. $x^4 - 4 x^3 (2) + 6 x^2 (2)^2 - 4 x (2)^3 + (2)^4$
    Write the skeleton with alternating signs, the powers of x running down from 4 and the powers of 2 running up.
  4. $2, 4, 8, 16$
    Evaluate the powers of 2 as their own pass — mixing that arithmetic with the sign pattern is how a minus goes missing.
  5. $4 \times 2 = 8$, $6 \times 4 = 24$, $4 \times 8 = 32$, $1 \times 16 = 16$
    Multiply term by term.
  6. $(x-2)^4 = x^4 - 8 x^3 + 24 x^2 - 32 x + 16$
    Combine the skeleton and the evaluated terms.
  7. 5 terms alternating from a plus; last term positive
    Check the signs — n is even, so the last term should be positive, and it is.
This is the worked case the chapter’s misconception check just ahead is built from: the sign alternates with r, a term’s position in the expansion, not with n, the exponent of the binomial — two numbers close enough together in the same expansion that it is easy to blur which one the sign actually follows.

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One count underneath every term

■RECAP

One combinatorial fact runs through the whole chapter.

In $(a+b)^n$, the term where r of the n brackets contribute a b has coefficient $C(n,r)$ — the count of ways to choose those r brackets. Everything else is bookkeeping on top of that.

PASCAL’S TRIANGLE is the same count read off a grid, row by row, using Pascal’s rule instead of the formula. Convenient for small n and whole rows.

THE GENERAL TERM, $T_{r+1} = C(n,r) a^{n-r} b^r$, is one summand pulled out so a single term can be found without the rest. Its index is offset by one from the term’s position, so $r = k-1$.

A MINUS SIGN changes only signs. Substituting $-b$ makes every odd-r term negative and leaves $C(n,r)$ alone.

THE MIDDLE TERM is the general term evaluated at the centre of the list — one term if n is even, two if n is odd, because the expansion has $n+1$ of them.

Every formula here is the same count with different bookkeeping. If one is forgotten, go back to choosing r brackets out of n.

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Two traps this chapter sets

✕MISCONCEPTION

THE TRAP. To expand $(x-2)^4$, work out $(x+2)^4$ and keep the answer as it stands, because the coefficients are the same either way: $x^4 + 8 x^3 + 24 x^2 + 32 x + 16$.

THE REALITY. Half of that is right, which is what makes it dangerous. The coefficients ARE the same. The SIGNS are not.

Substituting $-b$ leaves $C(n,r)$ untouched but attaches $(-1)^r$ to each term. When r is odd that factor is $-1$, so those terms turn negative.

The correct expansion is $x^4 - 8 x^3 + 24 x^2 - 32 x + 16$.

Compare the two term by term. The 1st, 3rd and 5th agree. The 2nd and 4th are off by a sign, and those are exactly the odd-r terms. Two terms out of five are wrong, and nothing about the arithmetic looks careless.

The check takes a moment. An $(a-b)^n$ expansion must alternate, starting with a plus. If two consecutive terms carry the same sign, something has been copied that should have been substituted.

The coefficients transfer between the two expansions. The signs never do.

Two consecutive powers on the sign rail alone: the alternation starts at a plus either way, and n decides only how far the rail runs and which sign it ends on.
✕MISCONCEPTION

THE TRAP. To find the 5th term of an expansion, put $r = 5$ into $C(n,r) a^{n-r} b^r$.

THE REALITY. That gives the 6th term. The formula is $T_{r+1} = C(n,r) a^{n-r} b^r$, so the subscript is $r+1$, not r.

Line the cases up and it is plain. $r = 0$ gives $T_1$. $r = 1$ gives $T_2$. $r = 2$ gives $T_3$. The value of r is always one less than the position.

So the 5th term needs $r + 1 = 5$, which is $r = 4$.

The offset is not arbitrary. r counts how many b factors a term carries, and the first term carries none — it is all a. Term positions start at 1 and r starts at 0, and the formula has to straddle both conventions.

What makes this one costly is that the wrong answer is a perfectly well-formed term. It has a sensible coefficient and sensible exponents. It is simply the wrong term, and nothing about it looks wrong.

The guard is the exponent. In the kth term the power of a should be $n - (k-1)$. Work that out independently and compare.

Subtract one from the position before substituting. Then check the exponent of a agrees.

Both lines are legal algebra and both return a real term — they are one column apart on the same ruler, which is why substituting r = 5 for the 5th term fails silently.

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Practice set

Exercise 7.1 — Expansion, coefficients, and the general term
  1. practice Expand $(x+3)^4$ using the binomial theorem.
  2. practice What is the coefficient of $a^3 b^4$ in the expansion of $(a+b)^7$?
    1. $21$
    2. $35$
    3. $42$
    4. $70$
  3. practice Find the 6th term in the expansion of $(x+2)^9$ using the general term formula.
  4. practice Find the term independent of x in the expansion of $(x + 1/x^2)^9$.
  5. practice Find the middle term in the expansion of $(x+3)^8$.
Answers
  1. $x^4 + 12 x^3 + 54 x^2 + 108 x + 81$
  2. $35$
  3. $4032 x^4$
  4. $84$
  5. $5670 x^4$
Exercise 7.1's three powers as term rails, each carrying the column its own question asks for: five terms for the fourth power, nine for the eighth, ten for the ninth.
Miscellaneous — practice set
  1. practice Expand $(x-2)^5$ using the binomial theorem.
  2. practice How many middle terms does the expansion of $(a+b)^9$ have?
    1. $1$
    2. $2$
    3. $3$
    4. $4$
Answers
  1. $x^5 - 10 x^4 + 40 x^3 - 80 x^2 + 80 x - 32$
  2. $2$
The eleven coefficients of (1+x) to the tenth drawn to scale: one peak at the middle term, and a mirror image either side of it.

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Chapter-end problems

Chapter-end problems — graded set
  1. board-easy Find the number of terms in the expansion of $(x+y)^9$.
  2. board-easy Using Pascal’s Triangle, write the coefficients in the expansion of $(a+b)^4$.
  3. board-easy Expand $(x+1)^3$ using the binomial theorem.
  4. board-easy Find the 3rd term, $T_3$, in the expansion of $(x+2)^5$.
    1. $40 x^3$
    2. $80 x^3$
    3. $20 x^3$
    4. $10 x^3$
  5. board-easy Find the 4th term in the expansion of $(a-b)^6$.
  6. board-easy Find $C(8,3)$, the coefficient of the $a^5 b^3$ term in $(a+b)^8$.
  7. board-easy How many terms does the expansion of $(2x-3y)^{10}$ have?
    1. $10$
    2. $11$
    3. $12$
    4. $9$
  8. board-easy Find the middle term of $(x+1)^4$.
  9. board-easy Find the middle terms of $(x+1)^5$.
  10. board-easy Expand $(x-2)^3$ using the alternating-sign expansion.
  11. board-standard Find the coefficient of $x^5$ in the expansion of $(x+3)^8$.
  12. board-standard Find the term independent of x in $(x + 1/x)^8$.
  13. board-standard Find the coefficient of $x^4$ in the expansion of $(2x-3)^6$.
  14. board-standard Find the 5th term in the expansion of $(x-2y)^7$.
  15. board-standard Find the middle term of $(2x+3y)^6$.
  16. board-standard Find the middle terms of $(x-1)^9$.
  17. board-standard Using Pascal’s Triangle, build row 6 from row 5 ($1, 5, 10, 10, 5, 1$) and read off the coefficients of $(a+b)^6$.
  18. board-standard Find the term independent of x in $(2x - 1/x)^6$.
Answers
  1. $10$ terms
  2. $1, 4, 6, 4, 1$
  3. $x^3+3x^2+3x+1$
  4. $40 x^3$
  5. $-20 a^3 b^3$
  6. $56$
  7. $11$
  8. $6x^2$
  9. $T_3=10x^3, T_4=10x^2$
  10. $x^3-6x^2+12x-8$
  11. $1512$
  12. $70$
  13. $2160$
  14. $560 x^3 y^4$
  15. $4320 x^3 y^3$
  16. $T_5=126x^5, T_6=-126x^4$
  17. $1, 6, 15, 20, 15, 6, 1$
  18. $-160$
When both parts of the binomial carry a number, the count is only one of three factors: the power of x rides with the 2 and r rides with the 3, so the x to the fourth term of (2x-3) to the sixth is 2160, not 15.
Chapter-end problems — graded set (continued)
  1. board-standard Find the coefficient of $x^3$ in the expansion of $(1-2x)^5$.
  2. board-standard Find the coefficient of $x^4$ in the expansion of $(3+2x)^5$.
  3. board-standard Find the coefficient of $x^7$ in the expansion of $(x+2)^{10}$.
  4. board-standard A binomial expansion has 9 terms. Find $n$.
    1. $8$
    2. $9$
    3. $7$
    4. $10$
  5. JEE Find the term independent of x in $(x^2 - 1/x)^9$.
    1. $84$
    2. $-84$
    3. $126$
    4. $36$
  6. JEE Find the coefficient of $x^5 y^7$ in the expansion of $(x+2y)^{12}$.
  7. JEE Find $n$ so that $C(14,n) = C(14,n+2)$.
  8. JEE Find the coefficient of $x^9$ in the expansion of $(x^2 - 1/x)^{12}$.
  9. JEE In the expansion of $(\sqrt{2} + \sqrt{3})^6$, how many terms are rational?
  10. JEE Find the sum of the binomial coefficients in the expansion of $(1+x)^{10}$.
    1. $1024$
    2. $100$
    3. $20$
    4. $512$
  11. JEE Find the middle term of $(2x-3y)^8$.
  12. JEE Find $r$ so that $C(37,r) = C(37,r+1)$.
Answers
  1. $-80$
  2. $240$
  3. $960$
  4. $n=8$
  5. $84$
  6. $101376$
  7. $n=6$
  8. $-792$
  9. $4$
  10. $1024$
  11. $90720 x^4 y^4$
  12. $r=18$
A term of (root 2 + root 3) to the sixth is rational only when both exponents are even, and because the power is even the two conditions shade the same columns: four of the seven terms survive.
JEE-tier practice — from the item bank
  1. Find the 5th term in the expansion of $(x+3)^7$.
    1. 945 $x^4$
    2. 2835 $x^3$
    3. 5103 $x^2$
    4. 81 $x^3$
    Check your answer
    1. 945 $x^4$ — C(7,3) $3^3$ $x^4$ = 945 $x^4$ uses r = 3, one step short of the r = 4 the 5th term needs.
    2. ✓ 2835 $x^3$ — (B) $T_5$ = $T_{r+1}$ with r=4 — C(7,4) $3^4$ $x^3$ = $35(81)x^3$ = 2835 $x^3.$
    3. 5103 $x^2$ — C(7,5) $3^5$ $x^2$ = 5103 $x^2$ uses r = 5, one step past the r = 4 the 5th term needs.
    4. 81 $x^3$ — 81 $x^3$ is just $3^4$ $x^3$ — the combination count C(7,4) was never applied.
  2. Find the 6th term in the expansion of $(x+2)^8$.
    1. 1792 $x^3$
    2. 1120 $x^4$
    3. 1792 $x^2$
    4. 32 $x^3$
    Check your answer
    1. ✓ 1792 $x^3$ — (A) $T_6$ = $T_{r+1}$ with r=5 — C(8,5) $2^5$ $x^3$ = $56(32)x^3$ = 1792 $x^3.$
    2. 1120 $x^4$ — C(8,4) $2^4$ $x^4$ = 1120 $x^4$ uses r = 4, one step short of the r = 5 the 6th term needs.
    3. 1792 $x^2$ — C(8,6) $2^6$ $x^2$ = 1792 $x^2$ uses r = 6 — the coefficient happens to match, but the power of x is wrong.
    4. 32 $x^3$ — 32 $x^3$ is just $2^5$ $x^3$ — the combination count C(8,5) was never applied.
  3. Find the coefficient of $x^4$ in the expansion of $(x+1)^9$.
    1. 84
    2. 126
    3. 210
    4. 36
    Check your answer
    1. 84 — C(9,3) = 84 chooses only 3 factors — the $x^4$ term needs exactly 4 chosen, not 3.
    2. ✓ 126 — (B) The $x^4$ term needs r = 4, choosing 4 factors of 1 from 9 — C(9,4) = 126.
    3. 210 — 210 is C(10,4) — one more than the actual n = 9 in this expansion, a different binomial entirely.
    4. 36 — 9 times 4 = 36 multiplies n and r instead of counting the ways to choose r of the n factors.
  4. Find the coefficient of $x^3$ in the expansion of $(2x+1)^7$.
    1. 560
    2. 84
    3. 280
    4. 35
    Check your answer
    1. 560 — C(7,3) $2^4$ = 560 uses r = 3, one step short of the r = 4 that gives $x^3.$
    2. 84 — C(7,5) $2^2$ = 84 uses r = 5, one step past the r = 4 that gives $x^3.$
    3. ✓ 280 — (C) $x^3$ needs r = 4 — C(7,4) $2^3$ = 35(8) = 280.
    4. 35 — C(7,4) = 35 drops the $2^3$ factor that $(2x)^{7-r}$ contributes.
  5. Find the middle term of $(x+2)^8$.
    1. 448 $x^5$
    2. 1792 $x^3$
    3. 70
    4. 1120 $x^4$
    Check your answer
    1. 448 $x^5$ — C(8,3) $2^3$ $x^5$ = 448 $x^5$ uses r = 3, one step short of the r = 4 the middle term needs.
    2. 1792 $x^3$ — C(8,5) $2^5$ $x^3$ = 1792 $x^3$ uses r = 5, one step past the r = 4 the middle term needs.
    3. 70 — 70 is just C(8,4) — the $2^4$ factor from $(2)^{8-r}$ was dropped.
    4. ✓ 1120 $x^4$ — (D) n=8 is even, one middle term at (8/2+1)=5th, r=4 — C(8,4) $2^4$ $x^4$ = $70(16)x^4$ = 1120 $x^4.$
Whether a row has one middle term or two is settled by nothing except the parity of n. An even n leaves a single centre chip; an odd n leaves none, so the two nearest share the job — and they always come out equal. Count the outlined boxes rather than reaching for a formula.
Subtracting a number, not just a letter: the minus 2 contributes a sign AND a size, so the x squared coefficient of (x-2) to the fifth is minus 80 rather than the count C(5,3) on its own.
JEE-tier practice — from the item bank (continued)
  1. Find the coefficient of the middle term of $(x+1)^{10}$.
    1. 210
    2. 252
    3. 200
    4. 120
    Check your answer
    1. 210 — C(10,4) = 210 uses r = 4 — the middle position is (10/2+1)=6th, which needs r = 5, not r = 4.
    2. ✓ 252 — (B) n=10 is even, one middle term at (10/2+1)=6th, r=5 — C(10,5) = 252.
    3. 200 — 200 does not match C(10,5); recompute the combination count directly.
    4. 120 — 120 is C(10,3), a different coefficient of this same row, not the middle one.
  2. Find the middle terms of $(x+1)^7$.
    1. 35 $x^4$ and 35 $x^3$
    2. 21 $x^5$ and 21 $x^2$
    3. 35 $x^4$ only
    4. 7 $x^6$ and 7 $x^1$
    Check your answer
    1. ✓ 35 $x^4$ and 35 $x^3$ — (A) n=7 is odd, two middle terms at the 4th and 5th, r=3,4 — C(7,3) $x^4$ = 35 $x^4$ and C(7,4) $x^3$ = 35 $x^3.$
    2. 21 $x^5$ and 21 $x^2$ — C(7,2) $x^5$ = 21 $x^5$ and C(7,5) $x^2$ = 21 $x^2$ use r = 2, 5 — one step off from the true r = 3, 4 on each side.
    3. 35 $x^4$ only — 35 $x^4$ alone treats n = 7 as if it were even, missing the second middle term entirely.
    4. 7 $x^6$ and 7 $x^1$ — C(7,1) $x^6$ = 7 $x^6$ and C(7,6) $x^1$ = 7 $x^1$ use r = 1, 6 — far from the true r = 3, 4.
  3. Find the middle terms of $(x+1)^9$.
    1. 462 $x^6$ and 462 $x^5$
    2. 126 $x^5$ only
    3. 126 $x^5$ and 126 $x^4$
    4. 36 $x^7$ and 36 $x^2$
    Check your answer
    1. 462 $x^6$ and 462 $x^5$ — C(11,5) $x^6$ = 462 $x^6$ and C(11,6) $x^5$ = 462 $x^5$ solve the middle terms of $(x+1)^{11}$, not the $(x+1)^9$ asked for.
    2. 126 $x^5$ only — 126 $x^5$ alone treats n = 9 as if it were even, missing the second middle term entirely.
    3. ✓ 126 $x^5$ and 126 $x^4$ — (C) n=9 is odd, two middle terms at the 5th and 6th, r=4,5 — C(9,4) $x^5$ = 126 $x^5$ and C(9,5) $x^4$ = 126 $x^4.$
    4. 36 $x^7$ and 36 $x^2$ — C(9,2) $x^7$ = 36 $x^7$ and C(9,7) $x^2$ = 36 $x^2$ use r = 2, 7 — far from the true r = 4, 5.
  4. Find the coefficient of $x^3$ in the expansion of $(x-1)^4$.
    1. 4
    2. -4
    3. 6
    4. -1
    Check your answer
    1. 4 — C(4,1) $x^3$ = 4 $x^3$ keeps the sign positive — r = 1 is odd, so this term must be negative.
    2. ✓ -4 — (B) $x^3$ needs r=1 (odd) — $C(4,1)(-1)^1$ = 4(-1) = -4.
    3. 6 — $C(4,2)(-1)^2$ $x^2$ = 6 $x^2$ uses r = 2, one step off from the r = 1 that gives $x^3.$
    4. -1 — -1 $x^3$ keeps only the sign, dropping the combination count C(4,1) entirely.
  5. Find the coefficient of $x^2$ in the expansion of $(x-2)^5$.
    1. 80
    2. 40
    3. 240
    4. -80
    Check your answer
    1. 80 — C(5,3) $2^3$ = 80 keeps the sign positive — r = 3 is odd, so this term must be negative.
    2. 40 — $C(5,2)(-2)^2$ = 40 uses r = 2, one step off from the r = 3 that gives $x^2.$
    3. 240 — $C(6,4)(-2)^4$ = 240 solves the $x^2$ term of $(x-2)^6$, not the $(x-2)^5$ asked for.
    4. ✓ -80 — (D) $x^2$ needs r=3 (odd) — $C(5,3)(-2)^3$ = 10(-8) = -80.
x⁴ needs 4 of the 9 factors, so the answer is C(9,4) = 126. Every wrong option is a real entry sitting next door: 84 is one chip left, 210 is the same column one row down, and 36 is r = 2. They are all correct answers to questions nobody asked.
The one thing the term count does not depend on is what the two parts are: coefficients and a minus sign change every term’s value, and all three rails still end at n + 1.
JEE-tier practice — from the item bank (continued)
  1. Find the number of terms in the expansion of $(x+y)^{12}$.
    1. 12
    2. 13
    3. 11
    4. 24
    Check your answer
    1. 12 — 12 is n itself — the expansion always has one MORE term than n, not n terms.
    2. ✓ 13 — (B) n=12, so the expansion has n+1 = 13 terms.
    3. 11 — 11 is two short of the true count — only one term is added to n, not subtracted.
    4. 24 — 24 doubles n by mistake — the term count adds 1 to n, it never multiplies it.
  2. Find the number of terms in the expansion of $(a-b)^{15}$.
    1. 16
    2. 15
    3. 14
    4. 30
    Check your answer
    1. ✓ 16 — (A) n=15, so the expansion has n+1 = 16 terms.
    2. 15 — 15 is n itself — the expansion always has one MORE term than n, not n terms.
    3. 14 — 14 is two short of the true count — only one term is added to n, not subtracted.
    4. 30 — 30 doubles n by mistake — the term count adds 1 to n, it never multiplies it.
  3. Using Pascal’s Triangle, find the coefficient of the 3rd term in the expansion of $(a+b)^6$.
    1. 10
    2. 20
    3. 15
    4. 21
    Check your answer
    1. 10 — 10 is C(5,2), row 5's entry — the question asks for row 6, the coefficients of $(a+b)^6$.
    2. 20 — 20 is C(6,3), one entry to the right of the 3rd term’s C(6,2).
    3. ✓ 15 — (C) The 3rd term uses r=2 — row 6's entry there is C(6,2) = 15.
    4. 21 — 21 does not match C(6,2); recompute the entry directly from the row.
  4. Find the coefficient of the $a^6 b^3$ term in the expansion of $(a+b)^9$.
    1. 126
    2. 36
    3. 84
    4. 27
    Check your answer
    1. 126 — C(9,4) = 126 chooses 4 factors — the $a^6$ $b^3$ term needs exactly 3 chosen, not 4.
    2. 36 — C(9,2) = 36 chooses only 2 factors — the $a^6$ $b^3$ term needs exactly 3 chosen, not 2.
    3. ✓ 84 — (C) The exponent of b is r = 3, so C(9,3) = 84.
    4. 27 — 9 times 3 = 27 multiplies n and r instead of counting the ways to choose r of the n factors.
  5. Find the term independent of x in the expansion of $(x + 1/x)^{10}$.
    1. 210
    2. 10
    3. 220
    4. 252
    Check your answer
    1. 210 — C(10,4) = 210 uses r = 4 — solving 10-2r=0 for r gives r = 5, not r = 4.
    2. 10 — 10 is just n itself — the exponent equation was never solved for r.
    3. 220 — 220 does not match C(10,5); recompute the combination count directly.
    4. ✓ 252 — (D) The general term gives $x^{10-2r}$. Setting 10-2r=0 gives r=5, so C(10,5)=252.
The term number counts from 1 and r counts from 0, so the 7th term carries r = 6 — one less, always. Its two powers are 5 and 6, and the bottom rail shows what makes that a check you can trust: the powers add to 11 in every column, not just this one.
The exponent ladder does not always fall by 2: with x squared in the first slot the step is 3, so the term independent of x sits at r = 6 and a reader who halves has stopped at the wrong rung.
JEE-tier practice — from the item bank (continued)
  1. Find the term independent of x in the expansion of $(x^2 + 1/x)^9$.
    1. 126
    2. 1
    3. 36
    4. 84
    Check your answer
    1. 126 — C(9,5) = 126 uses r = 5 — solving 18-3r=0 for r gives r = 6, not r = 5.
    2. 1 — C(9,9) = 1 comes from mis-solving the exponent equation as r = 9.
    3. 36 — C(9,7) = 36 uses r = 7 — one step past the r = 6 that zeroes the exponent.
    4. ✓ 84 — (D) The general term gives $x^{18-3r}$. Setting 18-3r=0 gives r=6, so C(9,6)=84.
  2. If the expansion of $(1+x)^n$ has 12 terms, find $n$.
    1. 12
    2. 11
    3. 10
    4. 13
    Check your answer
    1. 12 — 12 is the term count itself — n is always one LESS than the number of terms.
    2. ✓ 11 — (B) n+1 = 12, so n = 11.
    3. 10 — 10 subtracts two from the term count instead of one.
    4. 13 — 13 adds to the term count instead of subtracting from it.
  3. If $T_{r+1}$ is the 7th term of an expansion, find $r$.
    1. 7
    2. 6
    3. 5
    4. 8
    Check your answer
    1. 7 — r = 7 substitutes the term number straight in — but $T_{r+1}$ is the (r+1)th term, so r must be one LESS.
    2. ✓ 6 — (B) $T_{r+1}$ is the 7th term means r+1 = 7, so r = 6.
    3. 5 — r = 5 subtracts two from the term number instead of one.
    4. 8 — r = 8 adds to the term number instead of subtracting from it.
  4. In any single term of the expansion of $(a+b)^{11}$, what is the sum of the exponents of a and b?
    1. Always 11
    2. Varies from term to term
    3. Always 12
    4. Cannot be determined without r
    Check your answer
    1. ✓ Always 11 — (A) Every term is $a^{n-r} b^r$ — the exponents add to $(n-r)+r=n$, here 11, for every term.
    2. Varies from term to term — The individual exponents of a and b DO change from term to term, but their sum never does.
    3. Always 12 — 12 adds 1 to n by mistake — the exponent sum equals n exactly, nothing more.
    4. Cannot be determined without r — The sum is fixed at n for every term, whether or not r is known.
  5. Find the coefficient of $x^5$ in the expansion of $(x-3)^8$.
    1. 1512
    2. -1512
    3. 252
    4. -2268
    Check your answer
    1. 1512 — C(8,3) $3^3$ = 1512 keeps the sign positive — r = 3 is odd, so this term must be negative.
    2. ✓ -1512 — (B) $x^5$ needs r=3 (odd) — $C(8,3)(-3)^3$ = 56(-27) = -1512.
    3. 252 — $C(8,2)(-3)^2$ = 252 uses r = 2, one step off from the r = 3 that gives $x^5.$
    4. -2268 — $C(9,3)(-3)^3$ = -2268 solves the $x^5$ term of $(x-3)^9$, not the $(x-3)^8$ asked for.
An exam-tier coefficient question starts from a power of x, not a term number: find 5 on the middle rail, and both r and the count sit directly above and below it.
Ten columns for (x+y) to the ninth: four each side leaves two in the middle, so the term count is n + 1 and the middle is shared.

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