If S(x)=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+60(1+x)^60, x eq 0 , and (60)^2 ~S(60)=a(b)^b+b , where a, b \in N , then (a+b) equal to \underline\hspace2cm .

Sequences and Series PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q58 with Solution

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Correct answer: 3660

We need to evaluate $S(x)=(1+x)+2(1+x)^2+3(1+x)^3+ rac{}{ }\cdots+60(1+x)^{60}$ and then use the given relation $(60)^2S(60)=a(b)^b+b.$

Let $r=1+x.$ Then $S(x)=\sum_{k=1}^{60} k r^k.$

For the standard finite sum, $\sum_{k=1}^{n} k r^k=\frac{r\left(1-(n+1)r^n+nr^{n+1}\right)}{(1-r)^2}, \qquad r\neq 1.$

Here $n=60$ , so $S(x)=\frac{r\left(1-61r^{60}+60r^{61}\right)}{(1-r)^2}.$

Since $r=1+x$ , we have $1-r=1-(1+x)=-x,$ so $(1-r)^2=x^2.$ Hence $x^2S(x)=r\left(1-61r^{60}+60r^{61}\right).$

Now put $x=60$ . Then $r=1+60=61.$ Therefore $(60)^2S(60)=61\left(1-61\cdot 61^{60}+60\cdot 61^{61}\right).$

Factor powers of $61$ :

=61\left(1+59\cdot 61^{61}\right).$$ So $$(60)^2S(60)=61+59\cdot 61^{62}.$$ This is of the form $$a(b)^b+b.$$ Comparing, $$b=61, \qquad a=59\cdot 61.$$ Thus $$a=3599.$$ Therefore $$a+b=3599+61=3660.$$ ## Final Answer $$\boxed{3660}$$ The derived answer matches the stored correct answer.