Let $f(x)=\sum_{k=1}^{10} k x^{k}, x \in \mathbb{R}$. If $2 f(2)+f^{\prime}(2)=119(2)^{n}+1$ then $n

Question

Let $f(x)=\sum_{k=1}^{10} k x^{k}, x \in \mathbb{R}$. If $2 f(2)+f^{\prime}(2)=119(2)^{n}+1$ then $n$ is equal to

Professor PC Thomas Classes · Answer

(10): $f(x)=\sum_{k=1}^{10} k \cdot x^{k}, x \in R$\ $f(x)=x+2 x^{2}+\ldots .+10 x^{10}$\ $x \cdot f(x)=x^{2}+2 x^{3}+\ldots .+9 x^{10}+10 x^{11}$\ $f(x)(1-x)=x+x^{2}+x^{3}+\ldots .+x^{10}-10 x^{11}$\ $f(x)=\frac{x\left(1-x^{10}ight)}{(1-x)^{2}}-\frac{10 x^{11}}{(1-x)}$\ $f(x)=\frac{x-x^{11}-10 x^{11}(1-x)}{(1-x)^{2}}$\ $f(x)=\frac{x-x^{11}-10 x^{11}+10 x^{12}}{(1-x)^{2}}$ $$ =\frac{10 x^{12}-11 x^{11}+x}{(1-x)^{2}} $$ $$ (1-x)^{2} 	imes\left[120 x^{11}-121 x^{10}+1ight] $$ $f^{\prime}(x)=\frac{+2(1-x)\left[10 x^{12}-11 x^{11}+xight]}{(1-x)^{4}}$\ Hence, $2 f(2)+f^{\prime}(2)=119.2^{10}+1$\ $	herefore \quad n=10$\