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 ______.

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$$ I_k = \int_0^1 (1 - x)^k dx $$ $$ I_k = \left[(1 - x)^k \cdot x\right]_0^1 + \int_0^1 k(1 - x)^{k-1} \cdot (1 - x) dx $$ $$ I_k = 0 + k \int_0^1 (1 - x)^{k-1} dx - I_k $$ $$ I_k = -kI_k + kI_{k-1} $$ $$ I_k (1 + k) = kI_{k-1} $$ $$ \frac{I_k}{I_{k-1}} = \frac{k}{k + 1} $$ 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

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 ______.

Correct Answer: 65

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