Question

Let $r_k = \frac{\int_{0}^{1} (1 - x^7)^k \, dx}{\int_{0}^{1} (1 - x^7)^{k+1} \, dx}, \, k \in \mathbb{N}$. Then the value of $\sum_{k=1}^{10} \frac{1}{7(r_k - 1)}$is equal to ______.

Answer 1

$I_k = \int_0^1 (1 - x)^k dx$

$I_k = \left[(1 - x)^k \cdot x\right]_0^1 + \int_0^1 k(1 - x)^{k-1} \cdot (1 - x) dx$

$I_k = 0 + k \int_0^1 (1 - x)^{k-1} dx - I_k$

$I_k = -kI_k + kI_{k-1}$

$I_k (1 + k) = kI_{k-1}$

$\frac{I_k}{I_{k-1}} = \frac{k}{k + 1}$

Thus,

$r_k = \frac{7k + 8}{7k + 7}$

$r_k - 1 = \frac{-1}{7(k + 1)}$

Substituting in the summation:

$\sum_{k=1}^{10} \frac{1}{7(I_k - 1)} = \frac{1}{7} \cdot 7 \sum_{k=1}^{10} (k + 1) = \sum_{k=1}^{10} (k + 1)$

Computing:

$\sum_{k=1}^{10} (k + 1) = \sum_{k=1}^{10} k + \sum_{k=1}^{10} 1 = 55 + 10 = 65$

Final Answer: 65