Answer to Question #283308 in Algebra for pepeyo

Question

1+4+7...+(3n-2)=n(3n-1)/2

Answer 1

Let $P(n)$ denote the proposition that 

$$1+4+7...+(3n-2)=\dfrac{n(3n-1)}{2}$$

BASIS STEP:

$P(1)$ is true, because $1=\dfrac{1(3(1)-1)}{2}.$

INDUCTIVE STEP:

For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary

positive integer $k.$ That is, we assume that

$$1+4+7...+(3k-2)=\dfrac{k(3k-1)}{2}$$

Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that

$$1+4+7...+(3k-2)+(3(k+1)-2)$$

$$=\dfrac{(k+1)(3(k+1)-1)}{2}$$

When we add $(3(k+1)-2)$ to both sides of the equation in $P(k),$ we obtain

$$1+4+7...+(3k-2)+(3(k+1)-2)$$

$$=\dfrac{k(3k-1)}{2}+(3(k+1)-2)$$

$$=\dfrac{3k^2-k+6k+6-4}{2}$$

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

$$=\dfrac{(k+1)(3k+2)}{2}$$

$$=\dfrac{(k+1)(3(k+1)-1)}{2}$$

This last equation shows that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n.$ That is, we have proven that

$$1+4+7...+(3n-2)=\dfrac{n(3n-1)}{2}$$

for all positive integers $n.$