Verified Solution • 2025–26 Session moderate QJEEM24Ma1573: JEE Main 2024 Mathematics - Class 11 JEEM24Ma1573 Solution: Let α = ∑ k = 0 n ( ( n C k ) 2 k + 1 ) and β = ∑ k = 0 n − 1 ( … Let α=∑k=0n((nCk)2k+1) and β=∑k=0n−1(nCknCk+1k+2) If 5α=6β5 \alpha=6 \beta5α=6β , then nnn equals _______. Step-by-Step Verified Solution α=amp;∑k=0nnCk⋅nCkk+1⋅n+1n+1amp;=1n+1∑k=0nn+1Ck+1⋅nCn−kαamp;=1n+1⋅2n+1Cn+1βamp;=∑k=0n−1Ck⋅nCk+1k+2n+1n+1amp;1n+1∑k=0n−1nCn−k⋅n+1Ck+2amp;=1n+1⋅2n+1Cn+2βαamp;=2n+12n+1Cn+2βαamp;=2n+1−(n+2)+1n+2=56namp;=10\begin{aligned}\alpha= & \sum_{k=0}^n \frac{{ }^n C_k \cdot{ }^n C_k}{k+1} \cdot \frac{n+1}{n+1} \\& =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} \\\alpha & =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} \\\beta & =\sum_{k=0}^{n-1} C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} \\& \frac{1}{n+1} \sum_{k=0}^{n-1}{ }^n C_{n-k} \cdot{ }^{n+1} C_{k+2} \\& =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2} \\\frac{\beta}{\alpha} & =\frac{2 n+1}{2 n+1} C_{n+2} \\\frac{\beta}{\alpha} & =\frac{2 n+1-(n+2)+1}{n+2}=\frac{5}{6} \\n & =10\end{aligned}α=αβαβαβnamp;k=0∑nk+1nCk⋅nCk⋅n+1n+1amp;=n+11k=0∑nn+1Ck+1⋅nCn−kamp;=n+11⋅2n+1Cn+1amp;=k=0∑n−1Ck⋅k+2nCk+1n+1n+1amp;n+11k=0∑n−1nCn−k⋅n+1Ck+2amp;=n+11⋅2n+1Cn+2amp;=2n+12n+1Cn+2amp;=n+22n+1−(n+2)+1=65amp;=10 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