09092003, 05:51 PM  #1 (permalink) 
Upright

Discrete Math Help
Hello everyone, I was wondering if anyone here could help me out with a problem I'm trying to solve involving discrete math. The problem is an arithmaticgeometric series and I need to find the formula for the summation of it. Here it is:
The sum (sigma) from i=1 to n of [ i^2 * 3^i ] If someone could help me, or show me how to solve this, I'd greatly appreciate it, Thanks! 
09102003, 10:30 PM  #2 (permalink) 
Insane
Location: Bay Area

Ah I just had Discrete Math last Winter and I don't even remember how to do this stuff. I sold my book and I do not have my notes.
I think you can start about by breaking it into two pieces: for i = 1 to n sigma(i^2) * sigma(3^i) Then there should be some general rule for what to do with sigma(i^x) and sigma(x^i)... giving you a couple formulas to multiply. And that is what I would look up in my text book if I had it. Damn this is probably why I barely got a B in the class. 
09102003, 11:07 PM  #3 (permalink) 
Upright

Thanks for your suggestion. I was able to solve a solution for it, however it's quite lengthy. Here is my solution for anyone who might be interested in it :)
First, write out the terms of the series: x = 1*3^1 + 4*3^2 + 9*3^3 +...+ (n1)^2 * 3^(n1) + n^2 * 3^n Then, you multiply that by 3, to shift terms over by one: 3x = 1*3^2 + 4*3^3 +...+ (n2)^2 * 3^(n1) + (n1)^2 * 3^n + n^2 * 3^(n+1) Now subtract them: 3xx=2x= 1*3^1  3*3^2  5*3^3 ... (2n+3)*3^(n1) + (2n+1)*3^n + n^2 * 3^(n+1) so now we have what almost looks like a geometric series within a series... we must next form that geo series multiply by 3 again to shift terms once more: 2x*3=6x= 1*3^2  3*3^3 ... (2n+5)*3^(n1) + (2n+3)*3^n + (2n+1)*3^(n+1) + n^2 * 3^(n+2) next, subtract, 6x2x = 4x to show the geometric series within the series 4x = 1*3^1 + [2*3^2 + 2*3^3 +...+ 2*3^(n1) + 2*3^n] + (2nn^2+1)*3^(n+1) + n^2 * 3^(n+2) the series contained within the brackets is the geometric series we can sum up. Use the equation [ar^(n+1)  a] / r  1 2*3^(n+1)  18 / 3  1 Now that we have that, we apply that summation to the remaining 3 terms from the 4x= series.... Thus, 4x = [2*3^(n+1)  18]/2 + 3 + (2nn^2+1)*3^(n+1) + n^2*3^(n+2) Then get x by itself, so divide by 4: x = [(2*3^(n+1)  18)/2 + 3 + (2nn^2+1)*3^(n+1) + n^2*3^(n+2)] / 4 This is the final answer to find the sums of the first n terms of the series of i^2 * 3^i from i=1 to n. As you can see, when n=1, the first sum is 3, when n=2, the sum of the first two terms is 39, and when n=3, the sum of the first 3 terms is 282. It took awhile to derive this, but nonetheless this is a working solution. We have just solved an arithmaticgeometric series. :D 
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discrete, math 
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