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Please help - Series method of Differences

I am self studying maths, but I am really confused at the moment. I just don't know how to proceed.



For the first part, I need to come up with some expression for n^5, that expression involving differences.

How do I use the identity to accomplish this?

I will ask more questions about the rest of the question soon, I am just really confused about this first part.

Thanks buddies!
Original post by ybbob
I am self studying maths, but I am really confused at the moment. I just don't know how to proceed.



For the first part, I need to come up with some expression for n^5, that expression involving differences.

How do I use the identity to accomplish this?

I will ask more questions about the rest of the question soon, I am just really confused about this first part.

Thanks buddies!

Sum both sides of the identity from n=1 to N:
∑n=1N(n+12)6−(n−12)6=6∑n=1Nn5+5∑n=1Nn3+38∑n=1Nn\displaystyle\sum^N_{n=1}\left(n+\frac{1}{2}\right)^6 - \left(n-\frac{1}{2}\right)^6 = 6\displaystyle\sum^N_{n=1}n^5 +5\displaystyle\sum^N_{n=1}n^3 + \frac{3}{8}\displaystyle\sum^N_{n=1}n.

Using the method of differences, note that:
∑n=1N(n+12)6−(n−12)6=(N+12)6−(1−12)6\displaystyle\sum^N_{n=1}\left(n+\frac{1}{2}\right)^6 - \left(n-\frac{1}{2}\right)^6 = \left(N+\frac{1}{2}\right)^6 - \left(1-\frac{1}{2}\right)^6.

Therefore:
(N+12)6−126=6∑n=1Nn5+5∑n=1Nn3+38∑n=1Nn \left(N+\frac{1}{2}\right)^6 - \frac{1}{2^6} = 6\displaystyle\sum^N_{n=1}n^5 +5\displaystyle\sum^N_{n=1}n^3 + \frac{3}{8}\displaystyle\sum^N_{n=1}n

You should know what the following standard summations are:
∑n=1Nn3\displaystyle\sum^N_{n=1}n^3

∑n=1Nn\displaystyle\sum^N_{n=1}n

Can you see how you can find, from there:

∑n=1Nn5\displaystyle\sum^N_{n=1}n^5?
(edited 13 years ago)

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