Question

A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely.

Sn: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3

Answer 1

replace n with k and k+1

Answer 2

So Sk: 1 * 2 + 2 * 3 + 3 * 4 + . . .+ k(k + 1) = [k(k + 1)(k + 2)]/ 3 Would be the correct equation?

Answer 3

I think I'm a touch lost.

Answer 4

\[ \sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k = \frac{n (n+1) (2n+1)}{6} + \frac{n(n+1)}{2} \\ = \frac{n(n+1)}{2} \left( 1 + \frac{2n+1}{3}\right) \\
= \frac{n(n+1)(n+2)}{2}
\]

Answer 5

*
\[ \frac{n(n+1)(n+2)}{3} \]

Answer 6

I keep getting an [Math Processing Error] message. Sorry

Answer 7

this is a raw code
paste it in http://www.codecogs.com/latex/eqneditor.php

\sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k = \frac{n (n+1) (2n+1)}{6} + \frac{n(n+1)}{2} \\ = \frac{n(n+1)}{2} \left( 1 + \frac{2n+1}{3}\right) \\
= \frac{n(n+1)(n+2)}{3}

Answer 8

Thanks so much! That's going to be... fun writing out. -laughs-