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 $n$ equals ______.

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Binomial Theorem

Let

α = \sum_{k = 0}^{n} (\frac{( ^{n} C _{k} ) ^{2}}{k + 1})

and

β = \sum_{k = 0}^{n - 1} (\frac{^{n} C _{k} ^{n} C _{k + 1}}{k + 2})

. If

5 α = 6 β

, then

n

equals ______.

Answer: 10

Using

\frac{( k n )}{k + 1} = \frac{( k + 1 n + 1 )}{n + 1}

α = \sum_{k = 0}^{n} \frac{( k n ) ^{2}}{k + 1} = \frac{1}{n + 1} \sum_{k = 0}^{n} (k n) (k + 1 n + 1) = \frac{1}{n + 1} \sum_{k = 0}^{n} (n - k n) (k + 1 n + 1) = \frac{1}{n + 1} (n + 1 2 n + 1)

by Vandermonde's identity.

Using

\frac{( k + 1 n )}{k + 2} = \frac{( k + 2 n + 1 )}{n + 1}

β = \sum_{k = 0}^{n - 1} \frac{( k n ) ( k + 1 n )}{k + 2} = \frac{1}{n + 1} \sum_{k = 0}^{n - 1} (n - k n) (k + 2 n + 1) = \frac{1}{n + 1} (n + 2 2 n + 1)

.

So

\frac{β}{α} = \frac{( n + 2 2 n + 1 )}{( n + 1 2 n + 1 )} = \frac{2 n + 1 - ( n + 1 )}{( n + 1 ) + 1} = \frac{n}{n + 2}

.

Given

5 α = 6 β

\frac{β}{α} = \frac{5}{6}

, so

\frac{n}{n + 2} = \frac{5}{6} \Rightarrow 6 n = 5 n + 10 \Rightarrow n = 10

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Integer answer