# Question #c14ae

May 3, 2017

May 3, 2017

Assume true for $n = k$.

$\therefore {T}_{k} = 2 + 5 + 8 + \ldots + \left(3 k - 1\right) = \frac{1}{2} k \left(3 k + 1\right) q \quad \triangle$

And so it follows that:

${T}_{k + 1} = \frac{1}{2} k \left(3 k + 1\right) \textcolor{b l u e}{+ \left(3 \left(k + 1\right) - 1\right)}$

$= \frac{3}{2} {k}^{2} + \frac{1}{2} k + 3 k + 3 - 1$

$= \frac{3}{2} {k}^{2} + \frac{7}{2} k + 2$

$= \frac{1}{2} \left(3 {k}^{2} + 7 k + 4\right)$

Long divide by $k + 1$

$= \frac{1}{2} \left(k + 1\right) \left(3 k + 4\right)$

$= \frac{1}{2} \left(k + 1\right) \left(3 \left(k + 1\right) + 1\right)$

So if this is true for $n = k$, it is true for $n = k + 1 , k + 2 , k + 3 , \ldots$

Next, set $k = 1$ and we see that for $\star$, LHS = RHS = 2.

So it is true for $k = 1 , 2 , 3 , \ldots$, ie $k \in \mathbb{N}$