Chapter 9.4, Problem 47E

To determine

To find the sum ${\displaystyle \sum _{n=1}^6{n}^2}$ using the formulas for the sums of powers of integers.

Answer to Problem 47E

  ${\displaystyle \sum _{n=1}^6{n}^2}=91$

Explanation of Solution

Given:

Let $S={\displaystyle \sum _{n=1}^6{n}^2}$

Concept used:

  $1^2+2^2+3^2+4^2+\cdots +n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

Calculation:

Apply the formula for the sum of the squares of the first n positive integers, and obtain

  $\begin{array}{c}S={\displaystyle \sum _{n=1}^6{n}^2}\\ =1^2+2^2+3^2+4^2+5^2+6^2\\ =\frac{6\left(6+1\right)\left(2\times 6+1\right)}{6}\\ =7\times 13\\ =91\end{array}$