Combinatorics

**tarheelborn** · October 11th 2010, 01:15 PM

Sum the series $1^2+2^2+ \cdots+n^2$ by observing that $m^2 =2 \dbinom{m}{2} + \dbinom{m}{1}$ and using the identity $\dbinom{0}{k}+ \dbinom{1}{k} + \cdots + \dbinom{n}{k}= \dbinom{n+1}{k+1}$ .

**Jhevon** · October 11th 2010, 01:26 PM

Originally Posted by tarheelborn

Sum the series $1^2+2^2+ \cdots+n^2$ by observing that $m^2 =2 \dbinom{m}{2} + \dbinom{m}{1}$ and using the identity $\dbinom{0}{k}+ \dbinom{1}{k} + \cdots + \dbinom{n}{k}= \dbinom{n+1}{k+1}$ .

lets see what you've tried.

**tarheelborn** · October 11th 2010, 01:36 PM

I am really not sure where to start with this. I did verify that $m^2 = 2* \dbinom{m}{2}+ \dbinom{m}{1}$ . So $1^2 = 2* \dbinom{1}{2} + \dbinom{1}{1} = (2*0)+1 = 1$ , $2^2=2* \dbinom{2}{2}+ \dbinom{2}{1}=2(1)+2=4$ and $n^2=2* \dbinom{n}{2} + \dbinom{n}{1} = 2*\frac{n!}{2!(n-2)!} + \frac{n!}{1*(n-1)!}$ .

**Jhevon** · October 11th 2010, 01:56 PM

Originally Posted by tarheelborn

I am really not sure where to start with this. I did verify that $m^2 = 2* \dbinom{m}{2}+ \dbinom{m}{1}$ . So $1^2 = 2* \dbinom{1}{2} + \dbinom{1}{1} = (2*0)+1 = 1$ , $2^2=2* \dbinom{2}{2}+ \dbinom{2}{1}=2(1)+2=4$ and $n^2=2* \dbinom{n}{2} + \dbinom{n}{1} = 2*\frac{n!}{2!(n-2)!} + \frac{n!}{1*(n-1)!}$ .

you're simplifying prematurely. slow down. just plug in the expressions first:

$\displaystyle 1^2 + 2^2 + \cdots + n^2 = 2 {1 \choose 2} + {1 \choose 1} + 2 {2 \choose 2} + {2 \choose 1} + \cdots + 2 {n \choose 2} + {n \choose 1}$

Now what? (you can write out the expression for a few more terms if you don't see a pattern or what to do)

**tarheelborn** · October 11th 2010, 04:31 PM

I really just don't see it. It looks like it would be $\dbinom{n}{n+1} + \dbinom{n}{n}$ but that doesn't make sense.

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