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In one of the lecture notes I have, there is an exercise in which I have to prove the following equality with sums:

\begin{equation} \frac{\sum_{i=1}^{l} (2i)^{2}}{2} = \sum_{i=1}^{l} 2(l-i)^{2}. \end{equation}

My approach: \begin{align*} \sum_{i=1}^{l} (2i)^{2} &= 2 \left( \sum_{i=1}^{l} 2(l-i)^{2} \right) \\ \sum_{i=1}^{l} 4i^{2} &= 2 \left( \sum_{i=1}^{l} 2(l^{2} -2li + i^{2}) \right) \\ \sum_{i=1}^{l} 4i^{2} &= 2 \left( \sum_{i=1}^{l} 2l^{2} -4li + 2i^{2}) \right) \\ \sum_{i=1}^{l} 4i^{2} &= \sum_{i=1}^{l} 2(2(l^{2} -2li + i^{2})) \\ \sum_{i=1}^{l} 4i^{2} &= \sum_{i=1}^{l} 4l^{2} -8li + 4i^{2} \end{align*}

and from this point I have no idea how to continue-perhaps my approach is wrong.

Math123
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1 Answers1

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Firstly, the identity needs correcting - the sum should be from $i=0$. If you take the sum from $i=1$, the identity does not hold. For example, put $l=1$, then

\begin{equation} \frac{1}{2}\sum_{i=1}^{1}(2i)^{2} = 2 \end{equation}

whilst

\begin{equation} \sum_{i=1}^{1}2(l-i)^{2} = 0 \end{equation}

so no identity there! The original question obviously contains a typo.

What we need to actually prove (and which is an identity) is

\begin{equation} \frac{1}{2}\sum_{i=0}^{l}(2i)^{2} \equiv \sum_{i=0}^{l}2(l-i)^{2}. \end{equation}

We can take the factors of $2^{2}$ and $2$ outside the summations straight away, so we now have

\begin{equation} \sum_{i=0}^{l}i^{2} = \sum_{i=0}^{l}(l-i)^{2}. \end{equation}

Edit Actually, as @DS points out in the comments, this already proves the case, since every term in the sum of $i^{2}$ appears in the sum over $(l-i)^{2}$. However, for a longer proof, multiply out the brackets on the right

\begin{equation} \sum_{i=0}^{l}i^{2} = \sum_{i=0}^{l}(l^{2}-2li + i^{2}). \end{equation}

The first term in the brackets is a constant, so remembering that we are summing from $i=0$ to $l$, this will get multiplied by a factor of $(l+1)$. That is

\begin{equation} \sum_{i=0}^{l}l^{2} = (l+1)l^{2}. \end{equation}

The second term involves the summation over $i$. I shall assume that you are familiar with the identity

\begin{equation} \sum_{i=0}^{l}i = \frac{l(l+1)}{2}. \end{equation}

Inserting these results into the expression above, we now have

\begin{align} \sum_{i=0}^{l}i^{2} &= l^{2}(l+1) - 2l\frac{l(l+1)}{2} + \sum_{i=0}^{l}i^{2} \\ &= \sum_{i=0}^{l}i^{2}, \end{align}

as required.

Since we have now proved

\begin{equation} \frac{1}{2}\sum_{i=0}^{l}(2i)^{2} \equiv \sum_{i=0}^{l}2(l-i)^{2}, \end{equation}

it is easy to show explicitly why

\begin{equation} \frac{1}{2}\sum_{i=1}^{l}(2i)^{2} \not\equiv \sum_{i=1}^{l}2(l-i)^{2}, \end{equation}

since

\begin{equation} \frac{1}{2}\sum_{i=0}^{l}(2i)^{2} \equiv \frac{1}{2}\sum_{i=1}^{l}(2i)^{2} \end{equation}

but

\begin{align} \sum_{i=1}^{l}2(l-i)^{2} &\equiv \sum_{i=0}^{l}2(l-i)^{2} - 2l^{2} \\ &\not\equiv \sum_{i=0}^{l}2(l-i)^{2}. \end{align}

Hence the original expression is incorrect.

Martin Vaughan
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    $\sum_{i=0}^{l}i^{2} = \sum_{i=0}^{l}(l-i)^{2}$ is obvious reindexing. You have done too much after that. – D S Mar 09 '24 at 16:18
  • Not sure what you mean by that - the identity as written in the question is incorrect (there is obviously a typo). One has to put the sum from $i=0$ for the identity to hold. Thereafter, I have just shown how to prove it. What is your objection to that? – Martin Vaughan Mar 10 '24 at 11:40
  • First, I haven't downvoted, and I have no "objection" per se. I was just noting that the equation after "outside the summations straight away, so we now have" is just a reindexing of the sum, and everything you've done after that is a painfully long method. – D S Mar 10 '24 at 12:53
  • Fair enough! I missed the point you were making. – Martin Vaughan Mar 10 '24 at 13:57