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JEE Mains Maths · Binomial Theorem

Coefficient Sums by Substitution and Differentiation

Adding binomial coefficients, with or without weights: substitute a value of x into the expansion for plain and alternating sums, and pull out the weight r with the identity r·C(n, r) = n·C(n − 1, r − 1) for weighted sums.

All Binomial Theorem notesJEE Mains Maths strategy

Why this matters

Twenty-six PYQs, thirteen of them numerical answer, and 2025 alone has eight. Every sum here is one expansion evaluated at a chosen x — 1, −1, a complex root — or that expansion differentiated. Two ideas cover the page.

Concept 1 of 2: Substituting x = 1, −1 and other values

Write f(x)=∑arxrf(x)=\sum a_rx^r. Then f(1)f(1) is the sum of all coefficients, f(−1)f(-1) the alternating sum, and f(1)±f(−1)2\frac{f(1)\pm f(-1)}{2} the sums of even- and odd-indexed coefficients. For (1+x)n(1+x)^n this gives 2n2^n, 0 and 2n−12^{n-1}. Alternating sums of an odd set of coefficients, or x=ix=i and cube roots of unity, pick out other sub-sums.

Definition

  • ∑(nr)=2n\sum\binom nr=2^n; ∑(−1)r(nr)=0\sum(-1)^r\binom nr=0.
  • (n0)+(n2)+⋯=(n1)+(n3)+⋯=2n−1\binom n0+\binom n2+\dots=\binom n1+\binom n3+\dots=2^{n-1}.
  • Odd-indexed coefficients of ff: f(1)−f(−1)2\frac{f(1)-f(-1)}{2}.
  • Partial alternating sum: ∑r=0k(−1)r(nr)=(−1)k(n−1k)\sum_{r=0}^{k}(-1)^r\binom nr=(-1)^k\binom{n-1}{k}.

Even and odd parts

∑r evenar=f(1)+f(−1)2,∑r oddar=f(1)−f(−1)2\sum_{r\ \text{even}}a_r=\frac{f(1)+f(-1)}{2},\qquad\sum_{r\ \text{odd}}a_r=\frac{f(1)-f(-1)}{2}

Worked example

(1+x+x2)4=∑r=08arxr(1+x+x^2)^{4}=\sum_{r=0}^{8}a_rx^r. Find a0+a2+a4+a6+a8a_0+a_2+a_4+a_6+a_8.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 9 · Q63Moderate

Example 1 · Binomial Theorem · Coefficient Sums by Substitution and Differentiation

Let (1+x+2x2)20=a0+a1x+a2x2+…+a40x40\left( 1 +x+ 2x^{2} \right)^{20}=a_{0}+a_{1}x+a_{2}x^{2}+ \ldots +a_{40}x^{40}. then a1+a3+a5+…+a37a_{1}+a_{3}+a_{5}+ \ldots +a_{37} is equal to

Remove the terms outside the range

If the sum stops before the last odd coefficient (say at a37a_{37} out of a39a_{39}), compute the full odd sum and subtract the missing coefficient separately.
Drill 14 more on substituting x = 1, −1 and other values

Concept 2 of 2: Weighted sums: r C(n, r) and r² C(n, r)

The identity r(nr)=n(n−1r−1)r\binom nr=n\binom{n-1}{r-1} removes a factor rr, so ∑r(nr)=n2n−1\sum r\binom nr=n2^{n-1}. For r2r^2, write r2=r(r−1)+rr^2=r(r-1)+r and use r(r−1)(nr)=n(n−1)(n−2r−2)r(r-1)\binom nr=n(n-1)\binom{n-2}{r-2}. Equivalently, differentiate (1+x)n(1+x)^n and put x=1x=1. A weight like n−rn-r is nn minus rr, handled term by term.

Definition

  • r(nr)=n(n−1r−1)r\binom nr=n\binom{n-1}{r-1}.
  • ∑r(nr)=n2n−1\sum r\binom nr=n2^{n-1}.
  • ∑r(r−1)(nr)=n(n−1)2n−2\sum r(r-1)\binom nr=n(n-1)2^{n-2}.
  • ∑r2(nr)=n(n+1)2n−2\sum r^2\binom nr=n(n+1)2^{n-2}.

Weighted sums

∑rr(nr)=n 2n−1,∑rr2(nr)=n(n+1) 2n−2\sum_{r}r\binom nr=n\,2^{n-1},\qquad\sum_{r}r^2\binom nr=n(n+1)\,2^{n-2}

Worked example

Find ∑r=08(2r+1)(8r)\sum_{r=0}^{8}(2r+1)\binom8r.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q138Moderate

Example 2 · Binomial Theorem · Coefficient Sums by Substitution and Differentiation

If 12⋅( 15C1)+22⋅( 15C2)+32⋅( 15C3)1^{2}\cdot\left( \ ^{15}C_{1} \right)+2^{2}\cdot\left( \ ^{15}C_{2} \right)+3^{2}\cdot\left( \ ^{15}C_{3} \right) +….+152⋅( 15C15)=+ \ldots. +15^{2}\cdot\left( \ ^{15}C_{15} \right)= 2m⋅3n⋅5k2^{m}\cdot3^{n}\cdot5^{k}, where m,n,k∈Nm,n,k \in N, then m+n+km + n + k is equal to :-

r² is not r · r in the identity

Applying r(nr)=n(n−1r−1)r\binom nr=n\binom{n-1}{r-1} twice needs the second rr rewritten as (r−1)+1(r-1)+1. Treating r2(nr)r^2\binom nr as n2(n−2r−2)n^2\binom{n-2}{r-2} is the standard wrong turn.
Drill 10 more on weighted sums: r c(n, r) and r² c(n, r)

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)

  • Substituting x = 1, −1 and other values

    Even and odd parts

    ∑r evenar=f(1)+f(−1)2,∑r oddar=f(1)−f(−1)2\sum_{r\ \text{even}}a_r=\frac{f(1)+f(-1)}{2},\qquad\sum_{r\ \text{odd}}a_r=\frac{f(1)-f(-1)}{2}
  • Weighted sums: r C(n, r) and r² C(n, r)

    Weighted sums

    ∑rr(nr)=n 2n−1,∑rr2(nr)=n(n+1) 2n−2\sum_{r}r\binom nr=n\,2^{n-1},\qquad\sum_{r}r^2\binom nr=n(n+1)\,2^{n-2}

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.

Take the Binomial Theorem chapter testDrill all 26 questions