Let = ? k = 0 n ( ( C k n ) 2 k + 1 ) and = ? k = 0 n - 1 ( C k C k + 1 n n k + 2 ) . If 5 = 6 , then n equals ________ . [2024]

Question

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 ________ . [2024]

Smart Achievers · Accepted Answer

(10)

We have, $α = \sum_{k = 0}^{n} (\frac{{({}^{n}{C_{k}})}^{2}}{k + 1})$ and $β = \sum_{k = 0}^{n - 1} (\frac{{}^{n}{C_{k} {}^{n}{C_{k + 1}}}}{k + 2})$

Now, $α = \sum_{k = 0}^{n} \frac{{}^{n}{C_{k}}}{k + 1} \cdot {}^{n}{C_{k}}$

$= \sum_{k = 0}^{n} \frac{{}^{n + 1}{C_{k + 1}}}{n + 1} \cdot {}^{n}{C_{n - k}} = \frac{1}{n + 1} \sum_{k = 0}^{n} {}^{n + 1}{C_{k + 1} \cdot {}^{n}{C_{n - k}}}$

$= \frac{1}{n + 1} \cdot {}^{2 n + 1}{C_{n + 1}}$

Now, $β = \sum_{k = 0}^{n - 1} {}^{n}{C_{n - k} \cdot} \frac{n + 1}{(k + 2) (n + 1)} \cdot {}^{n}{C_{k + 1}}$

$= \frac{1}{n + 1} \sum_{k = 0}^{n - 1} {}^{n}{C_{n - k} \cdot {}^{n + 1}{C_{k + 2}}} = \frac{1}{n + 1} \cdot {}^{2 n + 1}{C_{n + 2}}$

$∴$ $\frac{β}{α} = \frac{{}^{2 n + 1}{C_{n + 2}}}{{}^{2 n + 1}{C_{n + 1}}} = \frac{(2 n + 1 - n - 1)! \cdot (n + 1)!}{(2 n + 1 - n - 2)! \cdot (n + 2)!}$

$= \frac{n! \cdot (n + 1)!}{(n - 1)! (n + 2) (n + 1)!} = \frac{n}{n + 2} = \frac{5}{6}$

$\Rightarrow n = 10$

Solution

Q.
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 ________ . [2024]

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Solution

Q. Let α=∑k=0n((Ckn)2k+1) and β=∑k=0n-1(Ck Ck+1nnk+2). If 5α=6β, then n equals ________ . [2024]

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Q.
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 ________ . [2024]