Chapter 12, Problem 28RE

To determine

To show: The given statement is true for all natural numbers , using the principle of mathematical induction .

Answer to Problem 28RE

the statement is true .

Explanation of Solution

Given information:

  $1^2+4^2+7^2+\mathrm{......}+{(3n-2)}^2=\frac{1}{2}n(6n^2-3n-1)$

Since,

Therefore substituting $n=1$ both sides

Left hand side: $1^2=1$

Right hand side ,

  $\begin{array}{l}=\frac{1}{2}(1)\left\{6{(1)}^2-3(1)-1\right\}\\ =\frac{1}{2}(6-3-1)\\ =\frac{1}{2}\times 2\\ =1\end{array}$

Therefore left hand side =right hand side

Hence, the statement is true .