2 + 5 + 8 + 11 + … + (3n − 1)
Question:

$2+5+8+11+\ldots+(3 n-1)=\frac{1}{2} n(3 n+1)$

Solution:

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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