Prove by the principle of mathematical induction that for all $n \in N$ :
$1 + 4 + 7 + . . . . + (3 n - 2) = \frac{1}{2} n (3 n - 1)$

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Solution

The statement S1 : 1 = $\frac{1}{2} (3 - 1) 2 = 1$ is true.
Assume that Sk : 1 + 4 + 7 + . . . + (3k − 2) = $\frac{k}{2} (3 k - 1)$ is true --(1)

and to prove that Sk+1 : 1 + 4 + 7 + . . . + (3(k + 1) − 2) = $\frac{k + 1}{2} (3 (k + 1) - 1)$ is true.
Observe that
LHS:
1 + 4 + 7 + . . . + (3(k + 1) − 2) = 1 + 4 + 7 + . . . + (3k +1)

= $\frac{k}{2} (3 k - 1) + (3 k + 1)$ from (1)
= $\frac{k * (3 k - 1) + 2 (3 k + 1)}{2}$

= $\frac{3 k^{2} - k + 6 k + 2}{2}$

= $\frac{3 k^{2} + 5 k + 2}{2}$

= $\frac{(k + 1) (3 k + 2)}{2}$ = $\frac{(k + 1) (3 (k + 1) - 1)}{2}$ =RHS
Thus by the Principle of Math Induction $S_{n}$ is true for all natural numbers n.

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