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JEE Mains Maths · Binomial Theorem

Sums with Fractions and Products of Coefficients

Sums where each coefficient is divided by r + 1, which integration or the identity C(n, r)/(r + 1) = C(n + 1, r + 1)/(n + 1) handles; and sums of products of two coefficients, which Vandermonde's identity turns into one coefficient.

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Why this matters

Seventeen PYQs, and five of them are from 2026. Seven divide by r + 1 and need one identity or one integral; ten multiply two coefficients together, and each is a single coefficient of a product of two expansions. Two ideas cover the page.

Concept 1 of 2: Coefficients divided by r + 1

Dividing by r+1r+1 is what integration does to xrx^r. So ∑(nr)r+1xr+1=∫0x(1+t)n dt\sum\frac{\binom nr}{r+1}x^{r+1}=\int_0^x(1+t)^n\,dt, and at x=1x=1 the sum is 2n+1−1n+1\frac{2^{n+1}-1}{n+1}. Without integrating, (nr)r+1=(n+1r+1)n+1\frac{\binom nr}{r+1}=\frac{\binom{n+1}{r+1}}{n+1} turns the sum into a plain sum of coefficients of (1+x)n+1(1+x)^{n+1}, missing the first term.

Definition

  • (nr)r+1=1n+1(n+1r+1)\frac{\binom nr}{r+1}=\frac{1}{n+1}\binom{n+1}{r+1}.
  • ∑r=0n(nr)r+1=2n+1−1n+1\sum_{r=0}^{n}\frac{\binom nr}{r+1}=\frac{2^{n+1}-1}{n+1}.
  • ∑r=0n(−1)r(nr)r+1=1n+1\sum_{r=0}^{n}\frac{(-1)^r\binom nr}{r+1}=\frac{1}{n+1}.
  • With powers of xx: integrate (1+x)n(1+x)^n from 0 to the value.

Absorption into n + 1

∑r=0n(nr)r+1=1n+1∑r=0n(n+1r+1)=2n+1−1n+1\sum_{r=0}^{n}\frac{\binom nr}{r+1}=\frac{1}{n+1}\sum_{r=0}^{n}\binom{n+1}{r+1}=\frac{2^{n+1}-1}{n+1}

Worked example

Find (50)+(51)2+(52)3+⋯+(55)6\binom50+\frac{\binom51}{2}+\frac{\binom52}{3}+\dots+\frac{\binom55}{6}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q89Moderate

Example 1 · Binomial Theorem · Sums with Fractions and Products of Coefficients

If  11C12+ 11C23+…..+ 11C910=nm\frac{\ ^{11}C_{1}}{2}+\frac{\ ^{11}C_{2}}{3}+ \ldots.. +\frac{\ ^{11}C_{9}}{10}=\frac{n}{m} with gcd(n,m)=1gcd(n,m) = 1, then n+mn + m is equal to

Trim the ends you are not given

If the sum stops at (119)10\frac{\binom{11}{9}}{10}, the identity gives 112∑(12j)\frac{1}{12}\sum\binom{12}{j} over j=2,…,10j=2,\dots,10 only. Subtract the missing (120),(121),(1211),(1212)\binom{12}{0},\binom{12}{1},\binom{12}{11},\binom{12}{12} from 2122^{12}.
Drill 6 more on coefficients divided by r + 1

Concept 2 of 2: Products of coefficients: Vandermonde

A sum ∑(mr)(nk−r)\sum\binom mr\binom n{k-r}, whose lower indices add to a constant kk, is the coefficient of xkx^k in (1+x)m(1+x)n(1+x)^m(1+x)^n, so it equals (m+nk)\binom{m+n}{k}. If the lower indices are equal instead, flip one with (nr)=(nn−r)\binom nr=\binom n{n-r} first. Weights r(nr)2r\binom nr^2 use the pairing trick: add the sum to itself written backwards.

Definition

  • Vandermonde: ∑r(mr)(nk−r)=(m+nk)\sum_r\binom mr\binom n{k-r}=\binom{m+n}k.
  • ∑r(nr)2=(2nn)\sum_r\binom nr^2=\binom{2n}n.
  • ∑r(nr)(nr+1)=(2nn+1)\sum_r\binom nr\binom n{r+1}=\binom{2n}{n+1}.
  • ∑rr(nr)2=n2(2nn)\sum_rr\binom nr^2=\frac n2\binom{2n}n (pair rr with n−rn-r).

Vandermonde's identity

∑r(mr)(nk−r)=(m+nk)\sum_{r}\binom mr\binom n{k-r}=\binom{m+n}{k}

Worked example

Find ∑r=05(5r)(7r)\sum_{r=0}^{5}\binom5r\binom7r.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q75Moderate

Example 2 · Binomial Theorem · Sums with Fractions and Products of Coefficients

The value of ∑r=022 22Cr 23Cr\sum_{r = 0}^{22} \ ^{22}C_{r}\ ^{23}C_{r} is

Match the indices before summing

∑(mr)(nr)\sum\binom mr\binom nr is not (m+nr)\binom{m+n}{r} — rr is the summation variable. Rewrite one factor so the lower indices add to a fixed number; that fixed number is the answer's lower index.
Drill 9 more on products of coefficients: vandermonde

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Coefficients divided by r + 1

    Absorption into n + 1

    ∑r=0n(nr)r+1=1n+1∑r=0n(n+1r+1)=2n+1−1n+1\sum_{r=0}^{n}\frac{\binom nr}{r+1}=\frac{1}{n+1}\sum_{r=0}^{n}\binom{n+1}{r+1}=\frac{2^{n+1}-1}{n+1}
  • Products of coefficients: Vandermonde

    Vandermonde's identity

    ∑r(mr)(nk−r)=(m+nk)\sum_{r}\binom mr\binom n{k-r}=\binom{m+n}{k}

Watch out for (2)

Test yourself on Binomial Theorem

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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