Let r k = 0 1 ( 1 - x 7 ) k ?d x 0 1 ( 1 - x 7 ) k + 1 ?d x , k ? N . Then the value of k = 1 10 1 7 ( r k - 1 ) is equal to _______ . [2024]

Question

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 _______ . [2024]

Smart Achievers · Accepted Answer

(65)

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

$= {[{(1 - x^{7})}^{k} x]}_{0}^{1} + 7 k \int_{0}^{1} {(1 - x^{7})}^{k - 1} x^{7} d x$

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

$\Rightarrow I_{k} (7 k + 1) = 7 k I_{k - 1} \Rightarrow I_{k + 1} (7 k + 8) = 7 (k + 1) I_{k}$

$\Rightarrow \frac{I_{k}}{I_{k + 1}} = \frac{7 k + 8}{7 k + 7} \Rightarrow r_{k} = \frac{7 k + 8}{7 k + 7}$

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

$∴ \sum_{k = 1}^{10} \frac{1}{7 (r_{k} - 1)} = \sum_{k = 1}^{10} (k + 1) = \frac{11 \times 12}{2} - 1 = 65$

Solution

Q.
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 _______ . [2024]

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Solution

Q. Let rk=∫01(1-x7)k dx∫01(1-x7)k+1 dx, k∈N. Then the value of ∑k=11017(rk-1) is equal to _______ . [2024]

Ans. (65) Let Ik=∫011(1-x7)kdx =[(1-x7)kx]01+7k∫01(1-x7)k-1x7dx ⇒-7k∫01(1-x7)k-1((1-x7)-1)dx=-7kIk+7kIk-1 ⇒Ik(7k+1)=7k Ik-1⇒Ik+1(7k+8)=7(k+1)Ik ⇒IkIk+1=7k+87k+7⇒rk=7k+87k+7 ⇒rk-1=17(k+1)⇒7(rk-1)=1k+1 ∴ ∑k=11017(rk-1)=∑k=110(k+1)=11×122-1=65

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Q.
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 _______ . [2024]