Let $r_k=\frac{\int_0^1\left(1-x^7\right)^k d x}{\int_0^1\left(1-x^7\right)^{k+1} d x}, k \in \mathbb{N}$.…
Let $r_k=\frac{\int_0^1\left(1-x^7\right)^k d x}{\int_0^1\left(1-x^7\right)^{k+1} d x}, k \in \mathbb{N}$. Then the value of $\sum_{k=1}^{10} \frac{1}{7\left(r_k-1\right)}$ is equal to________
Solution
$\begin{aligned} & I_K=\int 1 \cdot\left(1-x^7\right)^K d x \\ & I_K=\left.\left(1-x^7\right)^K x\right|_0 ^1+7 K \int_0^1\left(1-x^7\right)^{K-1} x^6 \cdot x d x \\ & I_K=-7 K \int_0^1\left(1-x^7\right)^{K-1}\left(\left(1-x^7\right)-1\right) d x \\ & I_K=-7 K I_K+7 K I_{K-1} \\ & \Rightarrow \frac{I_K}{I_{K+1}}=\frac{7 K+8}{7 K+7}\end{aligned}$ $\begin{aligned} & r_K=\frac{7 K+8}{7 \mathrm{~K}+7} \\ & \mathrm{r}_{\mathrm{K}}-1=\frac{1}{7(\mathrm{~K}+1)} \\ & \Rightarrow 7\left(\mathrm{r}_{\mathrm{K}}-1\right)=\frac{1}{\mathrm{~K}+1} \\ & \sum_{\mathrm{K}=1}^{10}(\mathrm{~K}+1)=11(6)-1=65\end{aligned}$