Chapter 11.5, Problem 37WE

To determine

To calculate:

Sum of given series by identifying pattern of series.

Answer to Problem 37WE

Sum of the ${\displaystyle \sum _{k=1}^{20}\left(2^k+k\right)}$ expressions is $2.097360\times {10}^6$ .

Explanation of Solution

Give information:

Expression:

  ${\displaystyle \sum _{k=1}^{20}\left(2^k+k\right)}$

Calculation:

It is given that,

  ${\displaystyle \sum _{k=1}^{20}\left(2^k+k\right)}$

So given series is neither arithmetic nor geometric.

To find the sum of given series formula used is,

  $={\displaystyle \sum _{k=1}^{20}\left(2^k+k\right)}$

  $=\left(2^1+1\right)+\left(2^2+2\right)+\left(2^3+3\right)+\left(2^4+4\right)+\left(2^5+5\right)+\left(2^6+6\right)+$

  $\left(2^7+7\right)+\left(2^8+8\right)+\left(2^9+9\right)+\left(2^{10}+10\right)+\left(2^{11}+11\right)+\left(2^{12}+12\right)$

  $+\left(2^{13}+13\right)+\left(2^{14}+14\right)+\left(2^{15}+15\right)+\left(2^{16}+16\right)+\left(2^{17}+17\right)+\left(2^{18}+18\right)+\left(2^{19}+19\right)+\left(2^{20}+20\right)$

  $=\left(2+1\right)+\left(4+2\right)+\left(8+3\right)+\left(16+4\right)+\left(32+5\right)+$

  $\left(64+6\right)+\left(128+7\right)+\left(256+8\right)+\left(512+9\right)+\left(1024+10\right)+\left(2048+11\right)$

  $+\left(4096+12\right)+\left(8192+13\right)+\left(16384+14\right)+\left(32768+15\right)+\left(65536+16\right)$

  $+\left(131072+17\right)+\left(262144+18\right)+\left(524288+19\right)+\left(1048576+20\right)$

  $=\left(3\right)+\left(6\right)+\left(11\right)+\left(20\right)+\left(37\right)+\left(70\right)+\left(135\right)+$

  $\left(264\right)+\left(521\right)+\left(1034\right)+\left(2059\right)+\left(4108\right)+\left(8205\right)+\left(16398\right)+\left(32783\right)$

  $+\left(65552\right)+\left(131089\right)+\left(262162\right)+\left(524307\right)+\left(1048596\right)$

  $=2.097360\times {10}^6$

Therefore, Sum of the ${\displaystyle \sum _{k=1}^{20}\left(2^k+k\right)}$ expressions is $2.097360\times {10}^6$ .