The sum of the binomial coefficients (( 50 C0/1)+( 50 C2/3)+( 50 C4/5)+ ldots+( 50 C50/51)) is equal to

Question

The sum of the binomial coefficients (( 50 C0/1)+( 50 C2/3)+( 50 C4/5)+ ldots+( 50 C50/51)) is equal to (A) (250/51) (B) (250 - 1/51) (C) (251 - 1/51) (D) (251 - 1/50)

Tardigrade Admin · Accepted Answer

(( 50 C0/1)+( 50 C2/3)+( 50 C4/5)+ ldots+(50 C50/51)) =(1/51)( 51 C1+ 51 C3+ 51 C5+ ldots) =(1/51) ⋅ 251-1=(250/51)

Q. The sum of the binomial coefficients $(\frac{^{50} C _{0}}{1} + \frac{^{50} C _{2}}{3} + \frac{^{50} C _{4}}{5} + \dots + \frac{^{50} C _{50}}{51})$ is equal to

$\frac{2 ^{50}}{51}$

$\frac{2 ^{50} - 1}{51}$

$\frac{2 ^{51} - 1}{51}$

$\frac{2 ^{51} - 1}{50}$

$(\frac{^{50} C _{0}}{1} + \frac{^{50} C _{2}}{3} + \frac{^{50} C _{4}}{5} + \dots + \frac{50 C _{50}}{51})$
$= \frac{1}{51} (^{51} C_{1} +^{51} C_{3} +^{51} C_{5} + \dots)$
$= \frac{1}{51} \cdot 2^{51 - 1} = \frac{2 ^{50}}{51}$

Q. The sum of the binomial coefficients (150C0​​+350C2​​+550C4​​+…+5150C50​​) is equal to

51250​

51250−1​

51251−1​

50251−1​