Verified Solution • 2025–26 Session moderate QJEEM24Madda0: JEE Main 2024 Mathematics - Class 12 JEEM24Madda0 Solution: Let . Then the value of ∑ k = 1 10 1 7 ( r k − 1 ) is equal to… Let rk=∫01(1−x7)kdx∫01(1−x7)k+1dx,k∈Nr_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}rk=∫01(1−x7)k+1dx∫01(1−x7)kdx,k∈N . Then the value of ∑k=11017(rk−1) is equal to _________. Step-by-Step Verified Solution rk=IaIb, where Ia=∫01(1−x7)kdxr_k=\frac{I_a}{I_b} \text {, where } I_a=\int_0^1\left(1-x^7\right)^k d xrk=IbIa, where Ia=∫01(1−x7)kdx amp;=(1−x7)k+1⋅x]01−∫01(k+1)(1−x7)k(−7x6)⋅xdxamp;=−7(k+1)∫01(1−x7)k(−1+1−x7)amp;Ib=−7(k+1)[−la+lb]amp;⇒rk=laIb=7k+87k+7=1+17(k+1)amp;17(rk−1)=(k+1)amp;⇒∑r=−11017(rK−1)=∑r=110(k+1)=11.122−1=65\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}amp;=(1−x7)k+1⋅x]01−∫01(k+1)(1−x7)k(−7x6)⋅xdxamp;=−7(k+1)∫01(1−x7)k(−1+1−x7)amp;Ib=−7(k+1)[−la+lb]amp;⇒rk=Ibla=7k+77k+8=1+7(k+1)1amp;7(rk−1)1=(k+1)amp;⇒r=−1∑107(rK−1)1=r=1∑10(k+1)=211.12−1=65 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
