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  4. The sum displaystyle ∑ r = 150 r⋅ (2r + 250 - r) is equal to

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

NTA AbhyasNTA Abhyas 2020Sequences and Series

Solution:

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