Chapter 11.4, Problem 35WE

To determine

To calculate: The value of ${\displaystyle \sum _{k=1}^6\left({\displaystyle \sum _{j=1}^{k}1}\right)}$

Answer to Problem 35WE

The value of ${\displaystyle \sum _{k=1}^6\left({\displaystyle \sum _{j=1}^{k}1}\right)}$ is 21.

Explanation of Solution

Given information: ${\displaystyle \sum _{k=1}^6\left({\displaystyle \sum _{j=1}^{k}1}\right)}$

Calculation:

  $\Rightarrow {\displaystyle \sum _{k=1}^6\left({\displaystyle \sum _{j=1}^{k}1}\right)}={\displaystyle \sum _{j=1}^{1}1}+{\displaystyle \sum _{j=1}^{2}1}+{\displaystyle \sum _{j=1}^{3}1}+{\displaystyle \sum _{j=1}^{4}1}+{\displaystyle \sum _{j=1}^{5}1}+{\displaystyle \sum _{j=1}^{6}1}$

  $\Rightarrow {\displaystyle \sum _{k=1}^6\left({\displaystyle \sum _{j=1}^k{\left(j\right)}^0}\right)}={\displaystyle \sum _{j=1}^1{\left(j\right)}^0}+{\displaystyle \sum _{j=1}^2{\left(j\right)}^0}+{\displaystyle \sum _{j=1}^3{\left(j\right)}^0}+{\displaystyle \sum _{j=1}^4{\left(j\right)}^0}+{\displaystyle \sum _{j=1}^5{\left(j\right)}^0}+{\displaystyle \sum _{j=1}^6{\left(j\right)}^0}$

  $\Rightarrow {\displaystyle \sum _{k=1}^6\left({\displaystyle \sum _{j=1}^k{\left(j\right)}^0}\right)}={\left(1\right)}^0+{\left(1\right)}^0+{\left(2\right)}^0+{\left(1\right)}^0+{\left(2\right)}^0+{\left(3\right)}^0+{\left(1\right)}^0+$

  ${\left(2\right)}^0+{\left(3\right)}^0+{\left(4\right)}^0+{\left(1\right)}^0+{\left(2\right)}^0+{\left(3\right)}^0+{\left(4\right)}^0+{\left(5\right)}^0+{\left(1\right)}^0+{\left(2\right)}^0+{\left(3\right)}^0+{\left(4\right)}^0+{\left(5\right)}^0+{\left(6\right)}^0$

  $\Rightarrow {\displaystyle \sum _{k=1}^6\left({\displaystyle \sum _{j=1}^k{\left(j\right)}^0}\right)}=21$