can any one solve this for me

[saudyonline]

dear all ,
i need to know how can i solve the following

Evaluate the sum

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thanks

[shmoe]

Hi, I'll give you a kick on the first one. Try breaking it up into a bunch of sums you know how to deal with. I'll "peel off" the first one for you:

\sum_{i=1}^{n}ia^{i}=\sum_{i=1}^{n}a^{i}+\sum_{i=2 }^{n}(i-1)a^{i}

The first sum is a geometric series starting with a, which you probably have a formula for. The second sum now has one less term. Do the same trick on the second sum, peel off a geometric series starting with a^2:

\sum_{i=1}^{n}a^{i}+\sum_{i=2}^{n}(i-1)a^{i}=\sum_{i=1}^{n}a^{i}+\sum_{i=2}^{n}a^{i}+\s um_{i=3}^{n}(i-2)a^{i}

So we have a geometric series starting with a, one starting with a^2, and our last sum now has one less term. Keep doing this to get a sum of geometric series starting at a, a^2, a^3, and so on. Use your formula for geometric series and life should be good.


This isn't the only way to do this sum of course. You could rewrite it as a\sum_{i=1}^{n}ia^{i-1}
and try to recognize the sum part as a derivative of a geometric series. You can use this to get a closed form (i.e. a 'nice' formula) for this sum. I'm not sure which way is preferable, I like double sums and I'm not sure our "nice" formla will be very nice, so I'd probably use the first.

[shmoe]

For the second one, you'll want to do a similar thing-get rid of the extra i factor and massage it into a sum of binomial coefficients which you should know a formula for their sum. Rewrite as:

\sum_{i=0}^{n}i\frac{n!}{i!(n-i)!}=\sum_{i=1}^{n}\frac{n!}{(i-1)!(n-i)!}

The i=0 term is just 0 so I dropped it. Now this isn't quite a sum of binomial coefficients, because the terms aren't binomial coefficients yet, the numerator of each term is off. I'll leave it to you to figure out how to correct this. Once you do, you should be able to recognize the remaining sum as a sum of certain binomial coefficients..