Induction Proof $2 + 5 + 8 + 11 + \cdots + (9n - 1) = \frac{3n(9n + 1)}{ 2}$

Question

I am looking for an induction proof...

$$2 + 5 + 8 + 11 + \cdots + (9n - 1) = \frac{3n(9n + 1)}{2}$$

when $n \geq 1$.

I know there are $3$ steps to this.
1) Check
2) Do $n = k$
3) Do $n = k + 1$

Problem is, I can't seem to get past the first step, when I check: $n = 1: (9(1) - 1) = 8$ but isn't it suppose to be come to $2$?

Answer 1

Let the sum be

$$S = 2 + 5 + 8 + 11 + ... + (9n - 4) + (9n - 1)$$

Let's add $S$ to itself

$$2S = [2 + (9n-1)] + [5 + (9n-4)] + ... + [(9n - 4) + 5] + [(9n - 1) + 2]$$

$$2S = [(9n+1)] + [(9n+1)] + ... + [(9n+1)] + [(9n + 1)]$$

So

$$2S = \left(\frac{(9n-1) - 2}{3} + 1\right)(9n+1)$$

$$S = \frac{3n(9n+1)}{2}$$

Answer 2

The correct form of the sum is :

$\sum_{k=1}^{n}(2+3(k-1))=2n+3\sum_{k=0}^{n-1}k=2n+3\dfrac{(n-1)n}{2}$

$=\dfrac{3n^{2}+n}{2}$. The "$9n-1$" form is incorrect!!

(I mean in the question and hence in the answers)!