Binomial Theorem
Summation Identities with Binomial Coefficients
nta_pyq_2024_jan
Grade 11
Question:
Let $\alpha = \displaystyle\sum_{k=0}^{n} \left(\frac{({}^nC_k)^2}{k+1}\right)$ and $\beta = \displaystyle\sum_{k=0}^{n-1} \left(\frac{{}^nC_k\cdot{}^nC_{k+1}}{k+2}\right)$. If $5\alpha = 6\beta$, then $n$ equals
Step-by-Step Solution
Key Concept: Use $\frac{{}^nC_k}{k+1}=\frac{1}{n+1}{}^{n+1}C_{k+1}$ and Vandermonde-type convolution to express $\alpha$ and $\beta$ in terms of central binomial coefficients. Then set $5\alpha=6\beta$.
$\alpha = \frac{1}{n+1}\sum_{k=0}^{n}{}^{n+1}C_{k+1}\cdot{}^nC_{n-k} = \frac{1}{n+1}{}^{2n+1}C_{n+1}$.
$\beta = \frac{1}{n+1}\sum_{k=0}^{n-1}{}^nC_{n-k}\cdot{}^{n+1}C_{k+2} = \frac{1}{n+1}{}^{2n+1}C_{n+2}$.
$\frac{\beta}{\alpha} = \frac{{}^{2n+1}C_{n+2}}{{}^{2n+1}C_{n+1}} = \frac{(2n+1)-(n+2)+1}{n+2} = \frac{n}{n+2}$.
$5\alpha=6\beta \Rightarrow \frac{\beta}{\alpha}=\frac{5}{6}=\frac{n}{n+2}\Rightarrow 5n+10=6n\Rightarrow n=10$.
Correct Answer: 10