NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

If for $n \in N$ , $\sum_{k=0}^{2n}(-1)^k \left[\binom{2n}{k}\right]^2 = A$ , then the value of $\sum_{k=0}^{2n}(-1)^k (k - 2n) \left[\binom{2n}{k}\right]^2$ is:

Choose the correct answer:

Explanation

1. Use the Property of Summation:

We can change the variable of summation by replacing $k$ with $(2n - k)$ . The limits remain $0$ to $2n$ .

Using the property $\binom{2n}{k} = \binom{2n}{2n-k}$ :

S = \sum_{k=0}^{2n} (-1)^{2n-k} (2n - k - 2n) \left[\binom{2n}{2n-k}\right]^2

S = \sum_{k=0}^{2n} (-1)^{2n} (-1)^{-k} (-k) \left[\binom{2n}{k}\right]^2

Since $2n$ is even, $(-1)^{2n} = 1$ and $(-1)^{-k} = (-1)^k$ :

S = \sum_{k=0}^{2n} (-1)^k (-k) \left[\binom{2n}{k}\right]^2

S = -\sum_{k=0}^{2n} (-1)^k k \left[\binom{2n}{k}\right]^2 \quad \dots \text{(Equation 1)}

2. Rewrite the original $S$ :

From the question:

S = \sum_{k=0}^{2n} (-1)^k k \left[\binom{2n}{k}\right]^2 - 2n \sum_{k=0}^{2n} (-1)^k \left[\binom{2n}{k}\right]^2

Given that $\sum_{k=0}^{2n} (-1)^k \left[\binom{2n}{k}\right]^2 = A$ , we have:

S = \sum_{k=0}^{2n} (-1)^k k \left[\binom{2n}{k}\right]^2 - 2nA \quad \dots \text{(Equation 2)}

3. Add Equation 1 and Equation 2:

S + S = \left[ -\sum_{k=0}^{2n} (-1)^k k \binom{2n}{k}^2 \right] + \left[ \sum_{k=0}^{2n} (-1)^k k \binom{2n}{k}^2 - 2nA \right]

2S = -2nA

S = -nA

The correct option is (b) $-nA$ .