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JEE Mathematics 2023 Question with Solution

If (30C1)2+2(30C2)2+3(30C2)2+…+30(30C30)2=α⋅60!(30!)2({}^{30}C_1)^2+ 2({}^{30}C_2)^2 + 3({}^{30}C_2)^2 + \ldots + 30({}^{30}C_{30})^2 = \frac{\alpha \cdot 60!}{(30!)^2}, then α\alpha is equal to:

  • A

    3030

  • B

    6060

  • C

    1515

  • D

    1010

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

S=0⋅(30C0)2+1⋅(30C1)2+2⋅(30C2)2+⋯+30⋅(30C30)2S = 0\cdot \left({}^{30}C_0\right)^2 + 1\cdot \left({}^{30}C_1\right)^2 + 2\cdot \left({}^{30}C_2\right)^2 + \cdots + 30\cdot \left({}^{30}C_{30}\right)^2

Find: α\alpha in

S=α⋅60!(30!)2S = \frac{\alpha \cdot 60!}{(30!)^2}

From the extracted working,

S=30(30C0)2+29(30C1)2+28(30C2)2+⋯+0(30C30)2S = 30\left({}^{30}C_0\right)^2 + 29\left({}^{30}C_1\right)^2 + 28\left({}^{30}C_2\right)^2 + \cdots + 0\left({}^{30}C_{30}\right)^2

Adding these two forms,

2S=30[(30C0)2+(30C1)2+⋯+(30C30)2]2S = 30\left[\left({}^{30}C_0\right)^2 + \left({}^{30}C_1\right)^2 + \cdots + \left({}^{30}C_{30}\right)^2\right]

Using

∑r=030(30Cr)2=60C30\sum_{r=0}^{30} \left({}^{30}C_r\right)^2 = {}^{60}C_{30}

we get

S=15⋅60C30S = 15\cdot {}^{60}C_{30}

Now,

60C30=60!(30!)2{}^{60}C_{30} = \frac{60!}{(30!)^2}

Therefore,

S=15⋅60!(30!)2S = \frac{15\cdot 60!}{(30!)^2}

Hence, α=15\alpha = 15. The correct option is C.

Identity-Based Comparison

Given:

S=∑k=030k((30k))2S = \sum_{k=0}^{30} k\left(\binom{30}{k}\right)^2

Find: α\alpha such that

S=α⋅60!(30!)2S = \alpha \cdot \frac{60!}{(30!)^2}

Using the identity stated in the solution,

∑k=0nk((nk))2=n(2n−1n−1)\sum_{k=0}^{n} k\left(\binom{n}{k}\right)^2 = n\binom{2n-1}{n-1}

Substituting n=30n=30,

S=30(5929)S = 30\binom{59}{29}

Now write this in factorial form:

S=30⋅59!29! 30!S = 30\cdot \frac{59!}{29!\,30!}

Also,

α⋅60!(30!)2=α⋅60⋅59!(30!)2\alpha \cdot \frac{60!}{(30!)^2} = \alpha \cdot \frac{60\cdot 59!}{(30!)^2}

Comparing both expressions and simplifying gives

α=15\alpha = 15

Therefore, the correct option is C.

Common mistakes

  • Using ∑(nr)2=22n\sum \binom{n}{r}^2 = 2^{2n} is incorrect. The correct identity is ∑r=0n(nr)2=(2nn)\sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n}. Always apply the central binomial identity here.

  • Missing the symmetry step between r(30r)2r\binom{30}{r}^2 and (30−r)(30r)2(30-r)\binom{30}{r}^2 leads to an incomplete simplification. Pair the terms carefully before adding the two sums.

  • Comparing with α⋅60!(30!)2\frac{\alpha\cdot 60!}{(30!)^2} without rewriting (6030)\binom{60}{30} in factorial form can cause coefficient errors. First convert (6030)\binom{60}{30} to 60!(30!)2\frac{60!}{(30!)^2} and then read off α\alpha.

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