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