JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 29)
If $$\mathrm{S}(x)=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+60(1+x)^{60}, x \neq 0$$, and $$(60)^2 \mathrm{~S}(60)=\mathrm{a}(\mathrm{b})^{\mathrm{b}}+\mathrm{b}$$, where $$a, b \in N$$, then $$(a+b)$$ equal to _________.
Answer
3660
Explanation
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$$\begin{aligned} & (-x)(S(x))=\frac{(1+x)\left[(1+x)^{60}-1\right]}{(1+x-1)}-60(1+x)^{61} \\ & (-x) S(x)=\frac{(1+x)\left[(1+x)^{60}-1\right]}{x}-60(1+x)^{61} \\ & x S(x)=60(1+x)^{61}-\frac{(1+x)\left[(1+x)^{60}-1\right]}{x} \end{aligned}$$
Multiplying $$x$$ on both side,
$$x^2 S(x)=60 x(1+x)^{61}-(1+x)\left[(1+x)^{60}-1\right]$$
Putting $$x=60$$
$$\begin{aligned} & (60)^2 S(60)=60 \times 60(61)^{61}-(61)\left[61^{60}-1\right] \\ & =60 \times 60(61)^{61}-(61) \cdot 61^{60}+61 \\ & =(61)^{61}[60 \times 60-1]+61 \\ & =(3600-1) \cdot 61^{61}+61 \\ & a=3600-1, \quad b=61 \Rightarrow a+b=3660 \end{aligned}$$
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