JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 28)

Let $$r_k=\frac{\int_0^1\left(1-x^7\right)^k d x}{\int_0^1\left(1-x^7\right)^{k+1} d x}, k \in \mathbb{N}$$. Then the value of $$\sum_\limits{k=1}^{10} \frac{1}{7\left(r_k-1\right)}$$ is equal to _________.
Answer
65

Explanation

$$r_k=\frac{I_a}{I_b} \text {, where } I_a=\int_0^1\left(1-x^7\right)^k d x$$

JEE Main 2024 (Online) 6th April Morning Shift Mathematics - Definite Integration Question 13 English Explanation

$$\begin{aligned} & \left.=\left(1-x^7\right)^{k+1} \cdot x\right]_0^1-\int_0^1(k+1)\left(1-x^7\right)^k\left(-7 x^6\right) \cdot x d x \\ & =-7(k+1) \int_0^1\left(1-x^7\right)^k\left(-1+1-x^7\right) \\ & I_b=-7(k+1)\left[-l_a+l_b\right] \\ & \Rightarrow r_k=\frac{l_a}{I_b}=\frac{7 k+8}{7 k+7}=1+\frac{1}{7(k+1)} \\ & \frac{1}{7\left(r_k-1\right)}=(k+1) \\ & \Rightarrow \sum_{r=-1}^{10} \frac{1}{7\left(r_K-1\right)}=\sum_{r=1}^{10}(k+1)=\frac{11.12}{2}-1=65 \end{aligned}$$

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