Question

Find the value of ${\sum }_{i=1}^{\infty }$ ${\sum }_{j=1}^{\infty }$ ${\sum }_{k=1}^{\infty }$ $\frac{1}{2^i{2}^j{2}^k}$.

• A. 4
• B. 8
• C. 16
• D. 27

Answer 1

The correct option is B

8

${\sum }_{i=1}^{\infty }$ ${\sum }_{j=1}^{\infty }$ ${\sum }_{k=1}^{\infty }$ $\frac{1}{2^i{2}^j{2}^k}$

Since, one summation operator uses only one variable

So,first find the summation for ${\sum }_{k=1}^{\infty }$ $\frac{1}{2^i{2}^j{2}^k}$ keeping i and j constant.

Similarly extend it for other summation part.

= ${\sum }_{i=1}^{\infty }$ ${\sum }_{j=1}^{\infty }$ $\frac{1}{2^i{2}^j}$ $(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+.....)$

= ${\sum }_{i=1}^{\infty }$ ${\sum }_{j=1}^{\infty }$ $\frac{1}{2^i{2}^j}$ $\left[\frac{1}{1-\frac{1}{2}}\right]$ = ${\sum }_{i=1}^{\infty }$ ${\sum }_{j=1}^{\infty }$ $\frac{2}{2^i{2}^j}$

= 2${\sum }_{i=1}^{\infty }$ $\frac{1}{2^i}$ $[\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+.....]$ = 2${\sum }_{i=1}^{\infty }$ $\frac{1}{2^i}$ $\left(\frac{1}{1-\frac{1}{2}}\right)$

= 4${\sum }_{i=1}^{\infty }$ $\frac{1}{2^i}$

= 4 $[\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+.....\infty ]$

= 4 $\times$ $\frac{1}{1-\frac{1}{2}}$

= 8