Let f(x)= _k=1^10kx^k,x R.

Question

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

Answer: 10

Accepted Answer

Given, $f(x)=\sum _{k=1}^{10}k{x}^{k}$

$\begin{gathered} f(x)=x+2{x}^{2}+\ldots \ldots \ldots +10{x}^{10} \ f(x).x={x}^{2}+2{x}^{3}+\ldots \ldots \ldots +9{x}^{10}+10{x}^{11} \ f(x)(1-x)=x+{x}^{2}+{x}^{3}+\ldots \ldots \ldots +{x}^{10}-10{x}^{11} \ 	herefore f(x)=\frac{x(1-{x}^{10})}{(1-x{)}^{2}}-\frac{10{x}^{11}}{(1-x)} \end{gathered}$

$\Rightarrow$ $f(x)=\frac{x-{x}^{11}-10{x}^{11}+10{x}^{12}}{(1-x{)}^{2}}=\frac{10{x}^{12}-11{x}^{11}+x}{(1-x{)}^{2}}$

So,

$\begin{gathered} f(2)=2(1-{2}^{10})+10\cdot {2}^{11} \ =2+18\cdot {2}^{10} \end{gathered}$

${f}^{'}(x)=\frac{{(1-x)}^{2}(120{x}^{11}-121{x}^{10}+1)+2(1-x)(10{x}^{12}-{11.x}^{11}+2)}{{(1-x)}^{4}}$

So,

${f}^{'}(x)=\frac{1({120.2}^{11}-{121.2}^{10}+1)+2(-1)({10.2}^{12}-{11.2}^{11}+2)}{{(-1)}^{4}}$

= ${2}^{10}(83)-3$

Hence $2f(2)+{f}^{'}(2)={119.2}^{10}+1$

$\Rightarrow n=10$

Solutions · 1