Let r_k= _1^0(1-x^7)^kdx _1^0(1-x^7)^k+1dx,k N.

Question

Let ${r}_{k}=\frac{{\int }_{1}^{0}{(1-{x}^{7})}^{k}dx}{{\int }_{1}^{0}{(1-{x}^{7})}^{k+1}dx},k\in N$. Then the value of $\sum _{k=1}^{10}\frac{1}{7({r}_{k}-1)}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 65

Accepted Answer

${r}_{k}=\frac{{I}_{a}}{{I}_{b}}{I}_{a}={\int }_{1}^{0}{(1-{x}^{7})}^{k}dx$

$\begin{gathered} {={(1-{x}^{7})}^{k+1}\cdot x]}_{1}^{0}-{\int }_{1}^{0}(k+1){(1-{x}^{7})}^{k}(-7{x}^{6})\cdot xdx \ =-7(k+1){\int }_{1}^{0}{(1-{x}^{7})}^{k}(-1+1-{x}^{7}) \ {I}_{b}=-7(k+1)[-{l}_{a}+{l}_{b}] \ \Rightarrow {r}_{k}=\frac{{l}_{a}}{{I}_{b}}=\frac{7k+8}{7k+7}=1+\frac{1}{7(k+1)} \ \frac{1}{7({r}_{k}-1)}=(k+1) \ \Rightarrow \sum _{r=-1}^{10}\frac{1}{7({r}_{K}-1)}=\sum _{r=1}^{10}(k+1)=\frac{11.12}{2}-1=65 \end{gathered}$

Solutions · 1