Question

$\sum _{k=1}^{\infty }\sum _{r=0}^k\frac{1}{3^k}{(}^k{C}_r)=$

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

$\sum _{k=1}^{\infty }\sum _{r=0}^k\frac{1}{3^k}{(}^k{C}_r)$ $=\sum _{k=1}^{\infty }\frac{1}{3^k}\sum _{r=0}^k{(}^k{C}_r)$ Since, $k$ does not depend on $r$.

We know that the sum of binomial coefficients is equal to $2^k$

or $\sum _{r=0}^k{(}_r^{k}C)=2^k$.

The equation is reduced to $\sum _{k=1}^{\infty }\frac{2^k}{3^k}$.

This is a geometric progression with $a=\frac{2}{3}$ and $r=\frac{2}{3}$

Since $r<1$, the sum of the infinite geometric series is given by $\frac{a}{1-r}$.

$Answer=\frac{\frac{2}{3}}{1-\frac{2}{3}}=\frac{\frac{2}{3}}{\frac{1}{3}}=2$