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 arithmatic-geometric 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! :cool:

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.

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 +...+ (n-1)^2 * 3^(n-1) + n^2 * 3^n

Then, you multiply that by 3, to shift terms over by one:
3x = 1*3^2 + 4*3^3 +...+ (n-2)^2 * 3^(n-1) + (n-1)^2 * 3^n + n^2 * 3^(n+1)

Now subtract them:
3x-x=2x= -1*3^1 - 3*3^2 - 5*3^3 -...- (-2n+3)*3^(n-1) + (-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^(n-1) + (-2n+3)*3^n + (-2n+1)*3^(n+1) + n^2 * 3^(n+2)

next, subtract, 6x-2x = 4x to show the geometric series within the series

4x = 1*3^1 + [2*3^2 + 2*3^3 +...+ 2*3^(n-1) + 2*3^n] + (-2n-n^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 + (-2n-n^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 + (-2n-n^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 arithmatic-geometric series. :-D

btw westothemax, if * is regular multiplication, then you can't break up the sum in such a way.