NIELIT 2021 Dec Scientist A - Section B: 59

Question

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

• A. $40 (2)^{40}$
• B. $40 (2)^{39}$
• C. $41 (2)^{40}$
• D. $41 (2)^{39}$

Answer 1

The summation is seems like ,

$1*\binom{41}{1}+3*\binom{41}{3}+5*\binom{41}{5}+.....+41*\binom{41}{41}$

this we need to find.

$(1+x)^{41}=1+\binom{41}{1}x+\binom{41}{2}x^{2}+.......+\binom{41}{41}x^{41}$

differentiating both side with respect to x, and putting x=1;

$41(1+1)^{40}=\binom{41}{1}+\binom{41}2{2}+\binom{41}{3}3+.......+\binom{41}{41}41$ ---------(1)

now,

$(1-x)^{41}=1-\binom{41}{1}x+\binom{41}{2}x^{2}-\binom{41}{3}x^{3}+.......-\binom{41}{41}41x^{41}$

differentiating both side with respect to x, and putting x=1;

$-41(1-1)^{40}=-\binom{41}{1}+\binom{41}{2}2-\binom{41}{3}3+.......-\binom{41}{41}41$------------(2)

by subtracting (1)-(2) we get,

$41*2^{40}=2*\binom{41}{1}+6\binom{41}{3}+10*\binom{41}{5}+14*\binom{41}{7}+.....+82*\binom{41}{41}$

$41*2^{40}=2*(1*\binom{41}{1}+3*\binom{41}{3}+5*\binom{41}{5}+7*\binom{41}{7}+.....+41*\binom{41}{41})$

$41*2^{39}=1*\binom{41}{1}+3*\binom{41}{3}+5*\binom{41}{5}+7*\binom{41}{7}+.....+41*\binom{41}{41}$

it is our required answer.

so correct option is option D.