Chapter 9.4, Problem 14E

To determine

To prove: The formula $1+4+7+10+\mathrm{....}+(3n-2)=\frac{n}{2}\left(3n-1\right)$ for every integer $n$ . Using mathematical induction.

Explanation of Solution

Given: $1+4+7+10+\mathrm{....}+(3n-2)=\frac{n}{2}\left(3n-1\right)$

The given formula, $1+4+7+10+\mathrm{....}+(3n-2)=\frac{n}{2}\left(3n-1\right)$

Let, $P\left(n\right)=1+4+7+10+\mathrm{....}+(3n-2)$

Put $n=1$

LHS: $P\left(1\right)=1$

RHS: $\frac{1}{2}\left(3-1\right)=1$

  $LHS=RHS=1$

  $P\left(n\right)$ is true for $n=1.$

So, $P\left(k\right)$ is true for $n=k$ where, $k\in Z^{+}$ .

  $\therefore 1+4+7+10+\mathrm{....}+(3k-2)=\frac{k}{2}\left(3k-1\right)\mathrm{....}(i)$

Now, prove for $P\left(k+1\right)$ .

Put $n=k+1$

For LHS,

  $\begin{array}{c}P\left(k+1\right)=1+4+7+10+\mathrm{....}+(3(k+1)-2)\\ =1+4+7+10+\mathrm{....}+3k+3-2\\ =1+4+7+10+\mathrm{....}+(3k-2)+3\\ =\frac{k}{2}(3k-1)+3\left[\text{Using}\left(i\right)\right]\\ =\frac{k+1}{2}(3k+2)\end{array}$

For RHS,

  $\begin{array}{l}=\frac{\left(k+1\right)}{2}\left(3(k+1)-1\right)\\ =\frac{\left(k+1\right)}{2}(3k+2)\end{array}$

  $LHS=RHS=\frac{\left(k+1\right)}{2}(3k+2)$

Hence, $P\left(k+1\right)$ is true.