JEE 2023 — Mathematics PYQ

Given $f(x)={\sum }_{k=1}^{10}k{x}^k$.
Let $S=x+2x^2+\cdots +10x^{10}$
This is an AGP. For $x=2$:
$f(2)=2+2(2^2)+3(2^3)+\cdots +10(2^{10})$
$f(2)=9\cdot 2^{11}+2$

Differentiating $f(x)$ and putting $x=2$:
$f^{\prime }(2)={\sum }_{k=1}^{10}{k}^2{2}^{k-1}$

Substituting in $2f(2)-f^{\prime }(2)$:
$2f(2)-f^{\prime }(2)=119(2{)}^{10}+1$

Comparing with $119(2{)}^n+1$:
$\Rightarrow n=10$