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← JEE Main 2014 · 12 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 73 of 90

JEE Main 2014 · 12 Apr · Shift 1, Mathematics question 73

If 1+x4+x5=∑i=05ai(1+x)i1+x^4+x^5=\displaystyle\sum_{i=0}^5a_i(1+x)^i, for all xx in R\mathbb R, then a2a_2 is :

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Answer and worked solution

Official answer: A

Put x=t−1x=t-1: 1+(t−1)4+(t−1)51+(t-1)^4+(t-1)^5. The coefficient of t2t^2 is
(42)+(52)(−1)3=6−10=−4.\binom42+\binom52(-1)^3=6-10=-4.

Topics:Binomial expansion
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