JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Definite Integrals

Let

r_{k} = \frac{\int _{0}^{1} ( 1 - x ^{7} ) ^{k} d x}{\int _{0}^{1} ( 1 - x ^{7} ) ^{k + 1} d x}

k \in N

. Then the value of

\sum_{k = 1}^{10} \frac{1}{7 ( r _{k} - 1 )}

is equal to ____.

Answer: 65

Solution

With

I_{k} = \int_{0}^{1} (1 - x^{7})^{k} d x

, integration by parts gives

I_{k} = \frac{7 k}{7 k + 1} I_{k - 1}

, so

r_{k} = \frac{I _{k}}{I _{k + 1}} = \frac{7 k + 8}{7 k + 7}

.
Then

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

, so

\frac{1}{7 ( r _{k} - 1 )} = k + 1

k = 1 \sum 10 (k + 1) = (1 + 2 + \dots + 10) + 10 = 55 + 10 = 65.

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