JEE Main 2023 — Sequences and Series Question with Solution

Given,

$f\left(x\right)=\sum _{k=1}^{10}k·x^k, x\in \mathrm{ℝ}$

$\Rightarrow f\left(x\right)=x+2x^2+3x^3+\dots \dots \dots +10x^{10}$

Now let $S=x+2x^2+3x^3+\dots \dots \dots +10x^{10}$

So, $S·x = x^2+2x^3+\dots \dots \dots \dots +9x^{10}+10x^{11}$

Now subtracting above equations we get,

$S\left(1-x\right)=x+x^2+x^3+\dots \dots +x^{10}-10x^{11}$

$\Rightarrow S\left(1-x\right)=\frac{x\left(1-x^{10}\right)}{1-x}-10x^{11}$

$\Rightarrow S=\frac{x\left(1-x^{10}\right)}{{\left(1-x\right)}^2}-\frac{10x^{11}}{\left(1-x\right)}=f\left(x\right)$

Now finding, $f\left(2\right)=2\left(1-2^{10}\right)+10·2^{11}$

$\Rightarrow f\left(2\right)=2+18·2^{10}$

Now finding $f^{\text{'}}\left(x\right)$ we get,

$f^{\text{'}}\left(x\right)=\frac{-10x^{11}}{{\left(1-x\right)}^2}-\frac{110x^{10}}{1-x}-\frac{10x^{10}}{{\left(1-x\right)}^2}+\frac{2x\left(1-x^{10}\right)}{{\left(1-x\right)}^3}+\frac{1-x^{10}}{{\left(1-x\right)}^2}$

$\Rightarrow f^{\text{'}}\left(2\right)=\frac{-10·2^{11}}{{\left(1-2\right)}^2}-\frac{110·2^{10}}{1-2}-\frac{10·2^{10}}{{\left(1-2\right)}^2}+\frac{2·2·\left(1-2^{10}\right)}{{\left(1-2\right)}^3}+\frac{1-2^{10}}{{\left(1-2\right)}^2}$

$\Rightarrow f^{\text{'}}\left(2\right)=-10·2^{11}+110·2^{10}-10·2^{10}-4\left(1-2^{10}\right)+1-2^{10}$

$\Rightarrow f^{\text{'}}\left(2\right)=83·2^{10}-3$

$2f\left(2\right)+f^{\text{'}}\left(2\right)=2\left(2+18·2^{10}\right)+83·2^{10}-3=119(2{)}^{10}+1$

$\therefore n=10$