Let C(n, 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 CS 5 Dec 2021 Official Paper

Answer 1

Concept:

We are given the expression:

\( \sum_{k=0}^{20}(2k+1) \cdot \binom{41}{2k+1} \)

This is the sum of products of odd-index binomial coefficients weighted by their index values.

We use the identity:

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

Splitting it into even and odd:

\( \sum_{\text{odd } r} r \cdot \binom{n}{r} = \sum_{\text{even } r} r \cdot \binom{n}{r} = \frac{n \cdot 2^{n-1}}{2} = n \cdot 2^{n-2} \)

For \( n = 41 \):

\( \sum_{k=0}^{20}(2k+1)\binom{41}{2k+1} = 41 \cdot 2^{39} \)