OK: so, I found that the hexagonal number for nth term is:

$\displaystyle 1+5+9+...+(4n-3) = n(2n-1)$

Proof:

(Eventually, I will want to get to $\displaystyle 1+5+...+(4(k+1)-3) = (k+1)(2(k+1)-1) = 2k^2 + 3k + 1$

SO:

Let P(n) be the statement $\displaystyle 1+5+...+(4n-3) = n(2n-1)$.

(Base case) Since $\displaystyle 4(1)-3 = 1$, which equals $\displaystyle (1)(2(1)-1)$, P(1) is true.

Let k be a pos. integer.

So $\displaystyle 1+5+...+(4k-3) = k(2k-1)$

Therefore, $\displaystyle 1+5+...+(4k-3) + (4(k+1) - 3) = k(2k-1) + (4(k+1) - 3)$

= $\displaystyle 2k^2 - k + 4k + 3$

= $\displaystyle 2k^2 + 3k + 3$

Did i do something wrong? I have an extra +2 in the mix.

Please work with what I have, if possible, I kinda need to use the Strong Induction.