Summation Problem

**jaclyn91** · September 7th 2009, 06:54 AM

Show that

Ʃn i2 = n(n+1)(2n+1)/6, n>=1
i=1

Should I use the formula
Ʃn i = n(n+1)/2
i=1

and if so, how would I go about doing this?

**Plato** · September 7th 2009, 07:00 AM

Use induction.

**Krizalid** · September 7th 2009, 07:49 AM

Originally Posted by jaclyn91

Should I use the formula
Ʃn i = n(n+1)/2
i=1

Yes, this works.

Now, use this simple fact: $i(i+1)=\frac{1}{3}\left( i(i+1)(i+2)-i(i-1)(i+1) \right).$

So the sum becomes $\sum\limits_{i=1}^{n}{i(i+1)}=\frac{1}{3}\sum\limi ts_{i=1}^{n}{\left( i(i+1)(i+2)-i(i-1)(i+1) \right)},$ now this is

$\underbrace{\sum\limits_{i=1}^{n}{i^{2}}}_{\text{Y ou need this}}+\underbrace{\sum\limits_{i=1}^{n}{i}}_{\tex t{You know this}}=\frac{1}{3}\underbrace{\sum\limits_{i=1}^{n }{\left( i(i+1)(i+2)-i(i-1)(i+1) \right)}}_{n(n+1)(n+2),\text{ because the above telescopes}}.$

I think you can take it from there.

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