Let \(C(n, \quad r)=\binom{n}{r}\). The value of \(\sum_{k=0}^{20}(2 k+1) C(41,2 k+1)\) is :

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NIELIT Scientific Assistant IT 5 Dec 2021 Official Paper

Answer 1

The correct answer is Option 1.

Key Points

We are given the summation: \(\sum_{k=0}^{20}(2k + 1)\cdot C(41, 2k + 1)\)
This involves binomial coefficients of odd indices from 1 to 41.
Using the identity: \(\sum_{r=0}^{n} r \cdot C(n, r) = n \cdot 2^{n-1}\)
Here, we apply a known result: \(\sum_{k=0}^{\lfloor n/2 \rfloor} (2k+1) \cdot C(n, 2k+1) = 2^{n-1} \cdot \frac{n}{2}\) for odd indexed binomial coefficients weighted by (2k+1).
For n = 41, this gives: \(\frac{41}{2} \cdot 2^{40} = 40 \cdot 2^{40}\)

Additional Information

Binomial summation identities are powerful tools in combinatorics.
This particular identity simplifies complex-looking weighted sums involving binomial coefficients.

Hence, the correct answer is: Option 1: 40 × 2⁴⁰