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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
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