elem. numb. theory - hexagonal number nth formula - proof

**cassiopeia1289** · September 14th 2008, 10:21 AM

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

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

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

SO:
Let P(n) be the statement $1+5+...+(4n-3) = n(2n-1)$ .
(Base case) Since $4(1)-3 = 1$ , which equals $(1)(2(1)-1)$ , P(1) is true.
Let k be a pos. integer.
So $1+5+...+(4k-3) = k(2k-1)$
Therefore, $1+5+...+(4k-3) + (4(k+1) - 3) = k(2k-1) + (4(k+1) - 3)$
= $2k^2 - k + 4k + 3$
= $2k^2 + 3k + 3$

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.

**o_O** · September 14th 2008, 10:28 AM

$k(2k-1) + 4(k+1) - 3$

$= 2k^2 - k + 4k + {\color{red}4} - 3$

$= 2k^2 + 3k + {\color{red}1}$

Just a little arithmetic error

**cassiopeia1289** · September 14th 2008, 11:14 AM

aw thank you! I always end up doing stupid stuff like that

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