Question

The sum $\sum _{r=1}^{50}r\cdot \left(2^r+2^{50-r}\right)$ is equal to

• A. $25\left(2^{50}-1\right)$
• B. $50\left(2^{50}-1\right)$
• C. $25\left(2^{51}-1\right)$
• D. $50\left(2^{51}-1\right)$

Answer 1

Answer: (D)

$\sum _{r=1}^{50}r\left(2^r+2^{50-r}\right)$
$=\left\{1\cdot 2+2\cdot 2^2+3\cdot 2^3+.\dots \dots ..+50\cdot 2^{50}\right\}+\left\{1\cdot 2^{49}+2\cdot 2^{48}+3\cdot 2^{47}+.\dots ..+50\cdot 2^0\right\}$
$=2^0\left(0+50\right)+2^1\left(1+49\right)+2^2\left(2+48\right)+.\dots \dots .+2^{48}\left(48+2\right)+2^{49}\left(49+1\right)+2^{50}\left(50+0\right)$
$=50\left\{1+2+2^2+.\dots ..+2^{49}+2^{50}\right]$
$=50\frac{\left(2^{51}-1\right)}{\left(2-1\right)}=50\left(2^{51}-1\right)$