Binomial Theorem Question 10
Question 10 - 2024 (30 Jan Shift 2)
Let $\alpha=\sum _{k=0}^{n}\left(\frac{\left({ }^{n} C _k\right)^{2}}{k+1}\right)$ and $\beta=\sum _{k=0}^{n-1}\left(\frac{{ }^{n} C _k{ }^{n} C _{k+1}}{k+2}\right)$
If $5 \alpha=6 \beta$, then $n$ equals
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Answer (10)
Solution
$$ \begin{aligned} \alpha & =\sum _{k=0}^{n} \frac{{ }^{n} C _k \cdot{ }^{n} C _k}{k+1} \cdot \frac{n+1}{n+1} \\ & =\frac{1}{n+1} \sum^{n+1}{ }^{n+1} C _{k+1} \cdot{ }^{n} C _{n-k} \\ \alpha & =\frac{1}{n+1} \cdot{ }^{2 n+1} C _{n+1} \\ \beta & =\sum _{k=0}^{n-1}{ }^{n} C _k \cdot \frac{{ }^{n} C _{k+1}}{k+2} \frac{n+1}{n+1} \\ & \frac{1}{n+1} \sum _{k=0}^{n-1} C _{n-k} \cdot{ }^{n+1} C _{k+2} \\ & =\frac{1}{n+1} \cdot{ }^{2 n+1} C _{n+2} \\ \frac{\beta}{\alpha} & =\frac{2 n+1}{2 n+1} C _{n+2} \\ \frac{\beta}{\alpha} & =\frac{2 n+1-(n+2)+1}{n+2}=\frac{5}{6} \end{aligned} $$
$$ n=10 $$
