JEE Maths›Binomial Theorem›Visual Solution
Visual SolutionPYQ 2024 · Jan 27 Shift 1Easy
Question
If AA denotes the sum of all the coefficients in the expansion of (1−3x+10x2)n(1 - 3x + 10x^2)^n and BB denotes the sum of all the coefficients in the expansion of (1+x2)n(1 + x^2)^n, then:
(A)A=B3A = B^3
(B)3A=B3A = B
(C)B=A3B = A^3
(D)A=3BA = 3B
Solution Path
Sum of coefficients == value at x=1x=1. A=8nA = 8^n, B=2nB = 2^n. Since 8=238 = 2^3, A=(2n)3=B3A = (2^n)^3 = B^3.
01Question Setup
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AA = sum of coefficients of (1−3x+10x2)n(1 - 3x + 10x^2)^n, BB = sum of coefficients of (1+x2)n(1 + x^2)^n. Find the relation.
Find relation between AA and BB
02Substitution Trick
2/4
Sum of coefficients of any polynomial = value at x=1x = 1. So A=(1−3+10)n=8nA = (1 - 3 + 10)^n = 8^n and B=(1+1)n=2nB = (1 + 1)^n = 2^n.
A=8n,  B=2nA = 8^n, \; B = 2^n
03Key ConnectionKEY INSIGHT
3/4
Since 8=238 = 2^3, we get A=8n=(23)n=(2n)3=B3A = 8^n = (2^3)^n = (2^n)^3 = B^3.
A=8n=(2n)3=B3A = 8^n = (2^n)^3 = B^3
04Final Answer
4/4
The relation between A and B is A=B3A = B^3.
A=B3\boxed{A = B^3} - Answer (A)
Concepts from this question1 concepts unlocked
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Sum of Coefficients via Substitution

EASY

Sum of all coefficients of f(x) = f(1). Sum of even-indexed = (f(1)+f(-1))/2

∑coefficients=f(1),∑even=f(1)+f(−1)2\sum \text{coefficients} = f(1), \quad \sum \text{even} = \frac{f(1)+f(-1)}{2}

Instant 30-second solve for coefficient sum questions. No expansion needed.

Coefficient problemsSum/difference of coefficientsSubstitution trick
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