JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 29)
Let $$\alpha=\sum_\limits{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$$ and $$\beta=\sum_\limits{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$$ If $$5 \alpha=6 \beta$$, then $$n$$ equals _______.
Answer
10
Explanation
$$\begin{aligned}
\alpha= & \sum_{k=0}^n \frac{{ }^n C_k \cdot{ }^n C_k}{k+1} \cdot \frac{n+1}{n+1} \\
& =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} \\
\alpha & =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} \\
\beta & =\sum_{k=0}^{n-1} C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} \\
& \frac{1}{n+1} \sum_{k=0}^{n-1}{ }^n C_{n-k} \cdot{ }^{n+1} C_{k+2} \\
& =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2} \\
\frac{\beta}{\alpha} & =\frac{2 n+1}{2 n+1} C_{n+2} \\
\frac{\beta}{\alpha} & =\frac{2 n+1-(n+2)+1}{n+2}=\frac{5}{6} \\
n & =10
\end{aligned}$$
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