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# Proof by induction

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1. Show that

can be written as

I have managed to prove the first summation but not the second here's what i've managed to do so far.

2. First, LaTeX issues: you need \sum_{r=1}^{2n} (and later \sum_{r-1}^{2n-1} or similar) to get the summations right.

As far as the actual maths goes, you seem very confused about what you're doing. I have no idea what you mean by "I have managed to prove the first summation"; you are not going to be able to "prove" one summation by itself - the point of the question is to replace the original sum by two "simpler" (in the sense that they are in the formula booklet) summations.

At this point I would suggest you set n = 3 (or 4), and explicitly write out the terms of the original sum and the terms of the two sums underneath. Hopefully this will show you what is actually going on here.
3. (Original post by DFranklin)
First, LaTeX issues: you need \sum_{r=1}^{2n} (and later \sum_{r-1}^{2n-1} or similar) to get the summations right.

As far as the actual maths goes, you seem very confused about what you're doing. I have no idea what you mean by "I have managed to prove the first summation"; you are not going to be able to "prove" one summation by itself - the point of the question is to replace the original sum by two "simpler" (in the sense that they are in the formula booklet) summations.

At this point I would suggest you set n = 3 (or 4), and explicitly write out the terms of the original sum and the terms of the two sums underneath. Hopefully this will show you what is actually going on here.
ok they're sorted.

I see so actually when n=3 we get 1³+2³+3³+4³+5³+6³-2³-4³-6³
But am i allowed to use the summations to work backwards and prove ti that way or does that not work?
4. Is it required that you have to use induction, or can you use evaluate each summation in terms of n and show that they are in fact the same.
5. (Original post by B_9710)
Is it required that you have to use induction, or can you use evaluate each summation in terms of n and show that they are in fact the same.
Here is the original question.

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