Question

The value of ${\sum }_{x=1}^{10}{\sum }_{r=0}^{r=x-1}(2^x-2^r)$ is

• A. $16392$
• B. $16398$
• C. $15462$
• D. $15468$

Answer 1

The correct option is B $16398$

$\sum _{x=1}^{10}\sum _{r=0}^{x-1}(2^x-2^r)$

Now, $\sum _{r=0}^{x-1}(2^x-2^r)=(2^x-2^1)+(2^x-2^2)+......+(2^x-2^{x-1})$

$=(2^x+2^x+2^x+........+xterms)-(2^0+2^1+2^2+......+2^{x-1})$

$=x{.2}^x-\left[\frac{2^0(2^x-1)}{2-1}\right]=x{.2}^x-(2^x-1)=x{.2}^x-2^x+1$

Now, $\sum _{x=1}^{10}\sum _{r=0}^{x-1}(2^c-2^r)=\sum _{x=1}^{10}x{.2}^x+\sum _{x=1}^{10}1$

$=\sum _{x=1}^{10}x{.2}^x-[2^1+2^2+2^3+........+2^{10}]+10$

$=\sum _{x=1}^{10}x{.2}^x-\left[\frac{2(2^{10}-1)}{2-1}\right]+10=\sum _{x=1}^{10}x{.2}^x-2^{11}+12$

$=(1.2+{2.2}^2+{3.2}^3+......+{10.2}^{10})-2^{11}+12$

$=18734-2048+12=16398$ Ans