Question

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

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

${\sum }_{r=0}^{r=k}\frac{1}{3^k}{}^k{C}_r$
$=\frac{1}{3^k}[1+{}^k{C}_1+{}^k{C}_2+....{}^k{C}_k]$
$=\frac{1}{3^k}{.2}^k$
$={\frac{2}{3}}^k$
Now
${\sum }_{r=1}^{r=k}{\frac{2}{3}}^k$
$={\frac{2}{3}}^1+{\frac{2}{3}}^2+{\frac{2}{3}}^3+{\frac{2}{3}}^4+...\infty$
This forms an infinite G.P.
Hence the sum of the G.P is
$=\frac{\frac{2}{3}}{1-\frac{2}{3}}$
$=2$