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


n
å
i=1 
 i2i

as a function of n?
Thanks.
By Olof Sisask on Sunday, September 02, 2001 - 07:26 pm:

Hiya,

Well here's one solution:



S= n
å
i=1 
 i.2i=2+2.22+3.23+¼+n.2n

2S=22+2.23+3.24+¼+n.2n+1
Now, let

x=2+22+23+24+¼+2n-n.2n+1=2.([2n-1]/[2-1])-n.2n+1=2n+1-2- n.2n+1=-(n-1).2n+1-2

Then 2S+x=S so that

S=-x=2+(n-1).2n+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.


n
å
r=1 
 xr=(xn+1-x)/(x-1)


Differentiating both sides, then multiplying by x

n
å
r=1 
 r xr = x([(n+1)xn-1]/(x-1)-(xn+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
n
å
r=0 
 rb

. 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!