Chapter 11.7, Problem 3CYP

To determine

To find a counterexample to disprove

  $1^2+2^2+3^2+\mathrm{....}+n^2=\frac{n\left(3n-1\right)}{2}$

Answer to Problem 3CYP

For $n=3$

Explanation of Solution

Given:

  $1^2+2^2+3^2+\mathrm{....}+n^2=\frac{n\left(3n-1\right)}{2}$

Calculation:

In order to find a counterexample to disprove

  $1^2+2^2+3^2+\mathrm{....}+n^2=\frac{n\left(3n-1\right)}{2}$

First, let’s check this is true for $n=1$ .

Left side, for $n=1$

  $1^2=1$

And right side:

  $\begin{array}{c}\frac{1\left(3\times 1-1\right)}{2}=\frac{1\left(3-1\right)}{2}\\ =\frac{2}{2}\\ =1\\ \text{=Left side}\end{array}$

Thus, formula is true for $n=1$ .

Similarly, for $n=2$ :

Left side, for $n=2$

  $\begin{array}{c}{1}^2+2^2=1+4\\ =5\end{array}$

And right side:

  $\begin{array}{c}\frac{2\left(3\times 2-1\right)}{2}=6-1\\ =5\\ \text{=Left side}\end{array}$

Thus, formula is true for $n=2$ .

Similarly, for $n=3$ :

Left side, for $n=3$

  $\begin{array}{c}{1}^2+2^2+3^2=1+4+9\\ =14\end{array}$

And right side:

  $\begin{array}{c}\frac{3\left(3\times 3-1\right)}{2}=\frac{3\left(9-1\right)}{2}\\ =\frac{3\times 8}{2}\\ =3\times 4\\ =12\\ \ne \text{Left side}\end{array}$

Thus, the given formula is not true for $n=3$ .

Thus, counter example to disprove the given formula is for $n=3$ .