$2+5+8+11+\ldots+(3 n-1)=\frac{1}{2} n(3 n+1)$
Let P(n) be the given statement.
Now,
$P(n)=2+5+8+\ldots+(3 n-1)=\frac{1}{2} n(3 n+1)$
Step 1:
$P(1)=2=\frac{1}{2} \times 1(3+1)$
Hence, $P(1)$ is true.
Step 2 :
Let $P(m)$ be true.
Then,
$2+5+8+\ldots+(3 m-1)=\frac{1}{2} m(3 m+1)$
To prove: $P(m+1)$ is true.
That is,
$2+5+8+\ldots+(3 m+2)=\frac{1}{2}(m+1)(3 m+4)$
$P(m)$ is equal to:
$2+5+8+\ldots+(3 m-1)=\frac{1}{2} m(3 m+1)$
Thus, we have :
$2+5+8+\ldots+(3 m-1)+(3 m+2)=\frac{1}{2} m(3 m+1)+(3 m+2)$
[Adding $(3 m+2)$ to both sides]
$\Rightarrow 2+5+8+\ldots+(3 m+2)=\frac{1}{2}\left(3 m^{2}+m+6 m+4\right)=\frac{1}{2}\left(3 m^{2}+7 m+4\right)$
$\Rightarrow 2+5+8+\ldots+(3 m+2)=\frac{1}{2}(3 m+4)(m+1)$
Thus, $P(m+1)$ is true.
By the $p$ rinciple of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.
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