Prove by induction

**Sylvie** · November 25th 2010, 01:40 AM

Hello!
Could somebody prove this for me?

1+3+6+...+n(n+1)/2=n(n+1)(n+2)/6

I mean for n=k+1

**emakarov** · November 25th 2010, 03:20 AM

See The Math Forum.

**tonio** · November 25th 2010, 03:20 AM

Originally Posted by Sylvie

Hello!
Could somebody prove this for me?

1+3+6+...+n(n+1)/2=n(n+1)(n+2)/6

I mean for n=k+1

What've you done and where're you stuck?

Tonio

**Sylvie** · November 25th 2010, 09:56 AM

Thank you, I think I understand ;]

**Archie Meade** · November 25th 2010, 10:32 AM

Originally Posted by Sylvie

Hello!
Could somebody prove this for me?

1+3+6+...+n(n+1)/2=n(n+1)(n+2)/6

I mean for n=k+1

P(k)

$\displaystyle\sum_{i=1}^k\frac{i(i+1)}{2}=\frac{k( k+1)(k+2)}{6}$

P(k+1)

$\displaystyle\sum_{i=1}^{k+1}\frac{i(i+1)}{2}=\fra c{(k+1)(k+2)(k+3)}{6}$

Proof that P(k+1) will also be true if P(k) is true

$\displaystyle\sum_{i=1}^{k+1}\frac{i(i+1)}{2}=\sum _{i=1}^k\frac{i(i+1)}{2}+\frac{(k+1)(k+2)}{2}=\sum _{i=1}^k\frac{i(i+1)}{2}+\frac{3(k+1)(k+2)}{6}$

which, if P(k) is valid, is

$\displaystyle\frac{k(k+1)(k+2)}{6}+\frac{3(k+1)(k+ 2)}{6}=\frac{(k+1)(k+2)}{6}(k+3)$

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