Mathematics · Sequence and Series

JEE Main 2024 — 6 April, Shift 2 — Question 23

If $S(x)=(1+x)+2(1+x)^{2}+3(1+x)^{3}+\ldots .$ . $+60(1+x)^{60}, x \neq 0$ , and

$(60)^{2} S(60)=a(b)^{b}+b$ , where $a, b \in N$ , then $(a+b)$ equal to $\qquad$

Answer: 3660

Numerical answer — enter this value.

$S(x)=(1+x)+2(1+x)^{2}+3(1+x)^{3}+. .+60(1+x)^{60}$

$(1+x) S=(1+x)^{2}+\ldots \ldots . \quad 59(1+x)^{60}+60(1+x)^{61}$

$-x S=\frac{(1+x)(1+x)^{60}-1}{x}-60(1+x)^{61}$

Put $x=60$

$-60 S=\frac{61\left((61)^{60}-1\right)}{60}-60(61)^{61}$

on solving $3660$

Answer key and solution verified before publishing.

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