(C0/1)+(C2/3)+(C4/5)+(C6/7)+........ =

Question

(C0/1)+(C2/3)+(C4/5)+(C6/7)+........ = (A) (2n+1/n+1) (B) (2n+1 -1/n+1) (C) (2n/n+1) (D) None of these

Tardigrade Admin · Accepted Answer

Putting the value of C0, C2, C4....., we get = 1+(n(n-1)/3.2!) + (n(n-1)(n-3)/5.4!)+ ..... = (1/n+1) [(n+1)+(n(n+1)n(n-1)/3!) + (n(n+1)n(n-1)n(n-2)(n-3)/5!)+.....] Put n + 1 = N = (1/N)[N+ (N(N-1)(N-2)/3!)+(N(N-1)(N-2)(N-3)(N-4)/5!)+.....] = (1/N) NC1 + NC3+NC5+..... = (1/N) 2N-1 = (2n/n+1) ∵ N = n+1

Q. $\frac{C _{0}}{1} + \frac{C _{2}}{3} + \frac{C _{4}}{5} + \frac{C _{6}}{7} + ........ =$

$\frac{2 ^{n + 1}}{n + 1}$

$\frac{2 ^{n + 1} - 1}{n + 1}$

$\frac{2 ^{n}}{n + 1}$

None of these

Q. 1C0​​+3C2​​+5C4​​+7C6​​+........=

n+12n+1​

n+12n+1−1​

n+12n​

None of these

Q. $\frac{C _{0}}{1} + \frac{C _{2}}{3} + \frac{C _{4}}{5} + \frac{C _{6}}{7} + ........ =$

$\frac{2 ^{n + 1}}{n + 1}$

$\frac{2 ^{n + 1} - 1}{n + 1}$

$\frac{2 ^{n}}{n + 1}$