Let = ^nCkk + 1 and = (^nCk ^nCk+1)k + 2. If 5 = 6, then n equals… - Solution | JEE Mathematics

NVAMediumJEE 2024Simple Applications

Let $\alpha = \sum \frac{{}^nC_k}{k + 1}$ and $\beta = \sum \frac{({}^nC_k \, {}^nC_{k+1})}{k + 2}$ . If $5\alpha = 6\beta$ , then $n$ equals:

Answer

Correct answer:10

Standard Method

Given:

\alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^2}{k+1}

and

\beta = \sum_{k=0}^{n-1} \frac{\binom{n}{k}\binom{n}{k+1}}{k+2}

with

5\alpha = 6\beta

Find: $n$

Use the identities

\frac{1}{k+1}\binom{n}{k} = \frac{1}{n+1}\binom{n+1}{k+1}

and

\frac{1}{k+2}\binom{n}{k+1} = \frac{1}{n+1}\binom{n+1}{k+2}

along with Vandermonde's identity.

For $\alpha$ ,

\alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^2}{k+1} = \sum_{k=0}^{n} \left(\frac{\binom{n}{k}}{k+1}\right)\binom{n}{k}

So,

\alpha = \frac{1}{n+1}\sum_{k=0}^{n} \binom{n}{k}\binom{n+1}{k+1}

Using symmetry,

\binom{n+1}{k+1} = \binom{n+1}{n-k}

Hence,

\sum_{k=0}^{n} \binom{n}{k}\binom{n+1}{n-k} = \binom{2n+1}{n}

Therefore,

\alpha = \frac{1}{n+1}\binom{2n+1}{n}

Detailed Algebra

For $\beta$ ,

\beta = \sum_{k=0}^{n-1} \frac{\binom{n}{k}\binom{n}{k+1}}{k+2} = \sum_{k=0}^{n-1} \binom{n}{k}\left(\frac{\binom{n}{k+1}}{k+2}\right)

Thus,

\beta = \frac{1}{n+1}\sum_{k=0}^{n-1} \binom{n}{k}\binom{n+1}{k+2}

Using symmetry,

\binom{n+1}{k+2} = \binom{n+1}{n-k-1}

So,

\sum_{k=0}^{n-1} \binom{n}{k}\binom{n+1}{n-k-1} = \binom{2n+1}{n-1}

Hence,

\beta = \frac{1}{n+1}\binom{2n+1}{n-1}

Ratio of Binomial Coefficients

Substitute the simplified forms into $5\alpha = 6\beta$ :

5\left(\frac{1}{n+1}\binom{2n+1}{n}\right) = 6\left(\frac{1}{n+1}\binom{2n+1}{n-1}\right)

Cancel $\frac{1}{n+1}$ :

5\binom{2n+1}{n} = 6\binom{2n+1}{n-1}

Now use

\frac{\binom{2n+1}{n}}{\binom{2n+1}{n-1}} = \frac{n+2}{n}

So,

5\cdot \frac{n+2}{n} = 6

which gives

5(n+2) = 6n

n = 10

Therefore, the value of $n$ is $10$ .

Using the wrong identity for $\frac{1}{k+1}\binom{n}{k}$ is a common mistake. It must be converted to $\frac{1}{n+1}\binom{n+1}{k+1}$ . Otherwise the Vandermonde form does not appear. Rewrite the term carefully before summing.
Applying Vandermonde's identity without using symmetry can lead to mismatched indices. Convert $\binom{n+1}{k+1}$ to $\binom{n+1}{n-k}$ and $\binom{n+1}{k+2}$ to $\binom{n+1}{n-k-1}$ first.
While solving $5\binom{2n+1}{n} = 6\binom{2n+1}{n-1}$ , students often expand factorials incorrectly. In particular, $n! = n(n-1)!$ and $(n+2)! = (n+2)(n+1)!$ must be used correctly before cancellation.

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