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