# Thread: Combinatorics - Sum of series

1. ## Combinatorics - Sum of series

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

2. Originally Posted by tarheelborn
Sum the series $\displaystyle 1^2+2^2+ \cdots+n^2$ by observing that $\displaystyle m^2 =2 \dbinom{m}{2} + \dbinom{m}{1}$ and using the identity $\displaystyle \dbinom{0}{k}+ \dbinom{1}{k} + \cdots + \dbinom{n}{k}= \dbinom{n+1}{k+1}$.
lets see what you've tried.

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

4. Originally Posted by tarheelborn
I am really not sure where to start with this. I did verify that $\displaystyle m^2 = 2* \dbinom{m}{2}+ \dbinom{m}{1}$. So $\displaystyle 1^2 = 2* \dbinom{1}{2} + \dbinom{1}{1} = (2*0)+1 = 1$, $\displaystyle 2^2=2* \dbinom{2}{2}+ \dbinom{2}{1}=2(1)+2=4$ and $\displaystyle n^2=2* \dbinom{n}{2} + \dbinom{n}{1} = 2*\frac{n!}{2!(n-2)!} + \frac{n!}{1*(n-1)!}$.
you're simplifying prematurely. slow down. just plug in the expressions first:

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

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

5. ## Combinatorics - sum of series

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