Induction - Summation

**nerdzor** · March 24th 2009, 01:52 AM

$3+6+9+...+3n=\frac{n}{2}(3n+3)$

**Showcase_22** · March 24th 2009, 02:26 AM

When n=1:

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

Hence true for n=1. Assume true for n=k:

$<br /> 3+6+9+...+3k=\frac{k}{2}(3k+3)<br />$

...and i'll leave it up to you to prove it for n=k+1!

**nerdzor** · March 24th 2009, 02:50 AM

Originally Posted by Showcase_22

...and i'll leave it up to you to prove it for n=k+1!

Thats where I'm stuck...

**stapel** · March 24th 2009, 04:12 AM

Originally Posted by nerdzor

Thats where I'm stuck...

What have you tried? How far have you gotten?

You plugged "k + 1" into the formula:

. . . . . $3\, +\, 6\, +\, 9\, +\, ...\, +\, 3k\, +\, 3(k\, +\, 1)$

You broke off the "just k" part so you could apply the assumption:

. . . . . $\left(3\, +\, 6\, +\, 9\, +\, ...\, +\, 3k\right)\, +\, 3k\, +\, 3$

. . . . . $\frac{k}{2}\left(3k\, +\, 3\right)\, +\, 3k\, +\, 3\, =\, \frac{k(3k\, +\, 3)}{2}\, +\, \frac{6k\, +\, 6}{2}$

. . . . . $=\, \frac{3k^2\, +\, 3k\, +\, 6k\, +\, 2}{2}\, =\, \frac{3k^2\, +\, 9k\, +\, 6}{2}$

. . . . . $=\, \frac{3(k^2\, +\, 3k\, +\, 2)}{2}\, =\, \frac{3(k\, +\, 2)(k\, +\, 1)}{2}$

. . . . . $=\, \frac{3(k\, +\, 1)((k\, +\, 1)\, +\, 1)}{2}$

Where did you go from here?

Please be complete. Thank you!

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