How do I get the equality $\sum_{l-s}^{l+s}(2j+1) = (2l+1)(2s+1)?$ I am having trouble evaluating $$\sum_{l-s}^{l+s}(2j+1)$$
I checked it for  $l=3$, and $s=1$, but I am looking for the general solution. My notes tell me that the expression should be equal to $$(2l+1)(2s+1)$$
but I don't see how that came about.
(I encountered this expression when I was looking at the number of possible states of total angular momentum of a coupled system.)
 A: HINT
$$
\sum_{j=l-s}^{l+s} (2j+1)
 = 2 \sum_{j=l-s}^{l+s} j + \sum_{j=l-s}^{l+s} 1
$$
and now the second sum is $2s$, while the first can be easily determined by noting that first and last terms add up to $2l$, so do second and next-to-last, etc.
A: You may know that the sum of the first $k$ odd numbers is $k^2$:
$$\sum_{j=0}^{k-1} \left(2j+1\right) = k^2$$
And then:
$$\sum_{j=l-s}^{l+s} \left(2j+1\right) = \sum_{j=0}^{l+s} \left(2j+1\right)-\sum_{j=0}^{l-s-1} \left(2j+1\right)=\left(l+s+1\right)^2-\left(l-s\right)^2=\ldots$$
And the desired result follows by $a^2-b^2=(a-b)(a+b)$.
A: $$\ sum_{l-s}^{l+s} (2j+1) = \sum_0^{l+s} (2j+1) - \sum_0^{l-s-1} (2j+1)$$
So you simply need to figure out a closed from of $\sum_0^x (2j+1)$.
$$ \sum_0^x (2j+1)= \sum_0^x (1) + 2 \cdot \sum_0^x (j) = x+1+x(x+1)=(x+1)^2$$
Thus plugging into the previous part:
$$ \sum_{l-s}^{l+s} (2j+1) = (l+s+1)^2-(l-s)^2 $$ 
A: This is due to a trick by Gauss. Note that
$$\sum^{l+s}_{j=l-s}(2j+1)=\sum^{s}_{j=-s}(2(l+j)+1)=\frac{1}{2}\sum^{s}_{j=-s}(2(l+j)+1)+(2(l-j)+1)=\frac{1}{2}\sum^{s}_{j=-s}(4l+2)=\frac{1}{2}(2s+1)(4l+2)=(2s+1)(2l+1).$$
A: Alternatively, note it is a sum of an arithmetic progression:
$$\sum_{l-s}^{l+s}(2j+1)=\underbrace{2(l-s)+1}_{a_1}+\underbrace{2(l-s+1)+1}_{a_2}+\cdots +\underbrace{2(l+s)+1}_{a_{2s+1}}=\\
=\frac{2(l-s)+1+2(l+s)+1}{2}\cdot (2s+1)=(2l+1)(2s+1).$$
Note: The number of terms is: $(l+s)-(l-s)+1=2s+1$ and the difference is $a_2-a_1=2$.
