NIELIT STQC STA 2021

Question

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

A)40(2)^40

B)40(2)^39

C)41(2)^40

D)41(2)^39

Comments

Can you mail, them the paper? we will add all the questions here.

Answer 1

The answer would be option D.

The theorems required for this are elegantly explained in the following youtube link

https://www.youtube.com/watch?v=OjvF7OcdNiQ

$∑^{20}_{k=0}$(2k + 1)C(41, 2k + 1)

= 41* $∑^{20}_{k=0}$C(40, 2k) 

Now the sum of Even terms would be equal to odd terms as per theorems in the youtube link hence each of those sums would be $2^{39}$

Hence the answer is 41 * $2^{39}$