Prove the following by the principle of mathematical induction: `2+5+8+11++(3n-1)=1/2n\ (3n+1)`

Explanation

Let P(n) be true for n=m, that is, we suppose that

`P(m)=2+5+8+11+...+(3m−1)= 1/2​ m (3m + 1)`

Now `P(m + 1) = P(m) + T_(m+1)`​

`=1/2​m(3m+1)+[3(m+1)−1]`

`=1/2​[3m^2+m+6m+6−2]`

`=1/2​[3m^2+7m+4]`

`=1/2​(m+1)(3m+4)`

`=1​/2(m+1)[3(m)+1]`

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