Chapter 11.5, Problem 38WE

To determine

To calculate:

Sum of given series by identifying pattern of series.

Answer to Problem 38WE

Sum of the ${\displaystyle \sum _{n=1}^{12}\left(2^n-1\right)}$ expressions is $8178$ .

Explanation of Solution

Give information:

Expression:

  ${\displaystyle \sum _{n=1}^{12}\left(2^n-1\right)}$

Calculation:

It is given that,

  ${\displaystyle \sum _{n=1}^{12}\left(2^n-1\right)}$

So given series is neither arithmetic nor geometric.

To find the sum of given series formula used is,

  $\begin{array}{l}={\displaystyle \sum _{n=1}^{12}\left(2^n-1\right)}\\ =\left(2^1-1\right)+\left(2^2-1\right)+\left(2^3-1\right)+\left(2^4-1\right)+\left(2^5-1\right)+\left(2^6-1\right)+\left(2^7-1\right)+\left(2^8-1\right)+\left(2^9-1\right)+\\ \left(2^{10}-1\right)+\left(2^{11}-1\right)+\left(2^{12}-1\right)\\ =\left(2-1\right)+\left(4-1\right)+\left(8-1\right)+\left(16-1\right)+\left(32-1\right)+\left(64-1\right)+\left(128-1\right)+\left(256-1\right)+\left(512-1\right)+\\ \left(1024-1\right)+\left(2048-1\right)+\left(4096-1\right)\\ =\left(1\right)+\left(3\right)+\left(7\right)+\left(15\right)+\left(31\right)+\left(63\right)+\left(127\right)+\left(255\right)+\left(511\right)+\left(1023\right)+\left(2047\right)+\left(4095\right)\\ =8178\end{array}$

Therefore, Sum of the ${\displaystyle \sum _{n=1}^{12}\left(2^n-1\right)}$ expressions is $8178$ .