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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.)

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    $\begingroup$ Can you do $\sum_{k=1}^n k$ ... if so, you should be able to write your sum using that. Another suggestion would be induction on $s$. $\endgroup$ – GEdgar Mar 12 at 13:39
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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.

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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 $a^2-b^2=(a-b)(a+b)$.

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This is due to a trick by Gauss. Note that $$\sum^{l+s}_{j=l-s}(2j+1)=\sum^{s}_{j=-s}(2(l+j)+1)=\frac{1}{2}\sum^{s}_{j=-s}(2(l+j)+1)+(2(l-j)+1)=\frac{1}{2}\sum^{s}_{j=-s}(4l+2)=\frac{1}{2}(2s+1)(4l+2)=(2s+1)(2l+1).$$

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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 $a_2-a_1=2$.

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$$\ sum_{l-s}^{l+s} (2j+1) = \sum_0^{l+s} (2j+1) - \sum_0^{l-s-1} (2j+1)$$

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:

$$ \sum_{l-s}^{l+s} (2j+1) = (l+s+1)^2-(l-s)^2 $$

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