Series Summing Methods

By Russel Dale on Sunday, September 02, 2001 - 12:23 am:

I have been trying to figure out this question: How can I express

\sum_{i = 1}^{n} i 2^{i}

as a function of n?
Thanks.

By Olof Sisask on Sunday, September 02, 2001 - 07:26 pm:

Hiya,

Well here's one solution:

S = \sum_{i = 1}^{n} i . 2^{i} = 2 + 2 . 2^{2} + 3 . 2^{3} + \dots + n . 2^{n}

$2 S = 2^{2} + 2 . 2^{3} + 3 . 2^{4} + \dots + n . 2^{n + 1}$
Now, let

$x = 2 + 2^{2} + 2^{3} + 2^{4} + \dots + 2^{n} - n . 2^{n + 1} = 2 . ([2^{n} - 1] / [2 - 1]) - n . 2^{n + 1} = 2^{n + 1} - 2 - n . 2^{n + 1} = - (n - 1) . 2^{n + 1} - 2$

Then $2 S + x = S$ so that

$S = - x = 2 + (n - 1) . 2^{n + 1}$

Regards,
Olof

By Brad Rodgers on Monday, September 03, 2001 - 05:48 am:

This proof using calculus actually just uses a very useful trick that's seen quite often.

\sum_{r = 1}^{n} x^{r} = (x^{n + 1} - x) / (x - 1)

Differentiating both sides, then multiplying by x

\sum_{r = 1}^{n} r x^{r} = x ([(n + 1) x^{n} - 1] / (x - 1) - (x^{n + 1} - x) / (x - 1)^{2})

By putting in x=2, one should get the answer given above.

It ends up that this trick can be used surprisingly often. An example not deviating far from above is finding

\sum_{r = 0}^{n} r^{b}

. Just use the same process above

b

times, then take the limit of when the

x = 1

. I'm not sure if a useful closed formula could come from this for all

b

, but at least for a given

b

it works. Try it for

b = 1

...
Brad

By Olof Sisask on Monday, September 03, 2001 - 11:29 am:

Nice Brad! That didn't occur to me at all!

Olof

By Russel Dale on Tuesday, September 04, 2001 - 11:26 pm:

Thanks both of you. I appreciate it a lot! Thanks!