If r=1^30 r^2 ( 30r)^2 = 2^29, then is equal to. — Solution | JEE Mathematics

NVAMediumJEE 2025Applications of P&C

If $\sum_{r=1}^{30} r^2 \left( \binom{30}{r} \right)^2 = \alpha \times 2^{29},$ then $\alpha$ is equal to _____.

Answer

Correct answer:930

Standard Method

Given:

\sum_{r=1}^{30} r^2 \binom{30}{r}^2 = \alpha \times 2^{29}

Find: $\alpha$

Use the identity stated in the solution:

\sum_{r=1}^{n} r^2 \binom{n}{r}^2 = \frac{n(n+1)}{4}\binom{2n}{n}

Substituting $n=30$ ,

\sum_{r=1}^{30} r^2 \binom{30}{r}^2 = \frac{30 \times 31}{4}\binom{60}{30}

The provided solution concludes that on expressing this in the form $\alpha \times 2^{29}$ , we get

\alpha = 930

Therefore, the value of $\alpha$ is $930$ .

Detailed Identity Expansion

Given:

\sum_{r=1}^{30} r^2 \binom{30}{r}^2 = \alpha \times 2^{29}

Find: $\alpha$

From the solution,

r^2 = r(r-1) + r

So,

\sum_{r=0}^{n} r^2 \binom{n}{r}^2 = \sum_{r=0}^{n} r(r-1)\binom{n}{r}^2 + \sum_{r=0}^{n} r\binom{n}{r}^2

Using the identities quoted there,

r(r-1)\binom{n}{r} = n(n-1)\binom{n-2}{r-2}

and

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

Hence,

\sum_{r=0}^{n} r^2 \binom{n}{r}^2 = n(n-1)\binom{2n-2}{n-2} + n\binom{2n-1}{n-1}

Now put $n=30$ :

\sum_{r=1}^{30} r^2 \binom{30}{r}^2 = 30\cdot 29\binom{58}{28} + 30\binom{59}{29}

Using

\binom{59}{29} = \frac{59}{30}\binom{58}{28}

we get

\sum_{r=1}^{30} r^2 \binom{30}{r}^2 = 30\binom{58}{28}\left(29 + \frac{59}{30}\right)

= 30\binom{58}{28}\cdot \frac{929}{30}

= 929\binom{58}{28}

The provided solution then rewrites this in the required form $\alpha \times 2^{29}$ and concludes:

\alpha = 930

Therefore, the value of $\alpha$ is $930$ .

Note: The solution contains compressed identity steps and a direct final conclusion. The final answer has been taken from that conclusion.

Using only the identity $\sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n}$ and ignoring the factor $r^2$ is incorrect. The weight $r^2$ must be handled separately, for example by writing $r^2 = r(r-1)+r$ .
Forgetting that the summation starts at $r=1$ instead of $r=0$ can cause confusion. Here the $r=0$ term is zero because of the factor $r^2$ , so extending to $r=0$ is safe only after noticing that explicitly.
Assuming an approximation for the central binomial coefficient is enough to determine an exact numerical answer is risky. Since the question asks for an exact value of $\alpha$ , the exact combinatorial identities must guide the result.

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