Let "r" be the positive real zero of P(x) = 9x^5 + 7x^2 + 9. The sum r^4 + 2r^9 +...........+ kr^(5k-1) can be represented as the rational number a/b in lowest terms. Find a + b.

Question

anonymous · Answer

@Hermeezey If the sum is a series that can be represented by the general term kr^(2k-1) then the other two terms don't fit for integral values of k.
4 = 2k-1 
k = 5/2
Is k any real number or is the error in the general term? 
A general form of kr^(5k-1) fits for the first two terms.

anonymous · Answer

I'm sorry it should have been:

kr^(5k-1). I made a mistake at first.

anonymous · Answer

okay, working on the solution now. Will be back as soon as its done.

experimentx · Answer

Let $ 5k -1 = y => k = {y+1 \over 5} $
$$ {y+1 \over 5} r^{y}  = {1 \over 5} \left ((y+1) r^y  \right ) = 1/5 {d \over dk}r^{y+1}$$
sum the geometric sequence inside d/dk and differentiate it. or just google for arithmetic-geometric sequence.

anonymous · Answer

@experimentX fuc#in brilliant.

experimentx · Answer

well nice job simplifying it too.

experimentx · Answer

Oops ... there should be been d/dy instead of d/dk

anonymous · Answer

i'm sorry could you help walk me through the summing part. I'm not quite sure if I should use the geometric series formula a/(1-r).

anonymous · Answer

I guess what I'm trying to say is could you take this one all the way home because I honestly have no idea where to go.

experimentx · Answer

a ( r^y - 1)/(r - 1) ... just differentiate this.

experimentx · Answer

you can do this without differentiating too http://mathforum.org/library/drmath/view/66996.html

anonymous · Answer

Wait I thought "r" was a positive zero of P(x). Wouldn't that imply that "r" is a numerical value and not a variable, which would also imply that d/dy r^(y+1) = ln(y+1)*r^y+1 and not (y+1)*r^y

anonymous · Answer

What our fellow thinker @experimentX  has done here is to modify the series of summation, we still have to find the root, which is still evading me, but he went ahead and turned the summation into an elegant and tamed creature. 
The trick is to use a fresh variable y, such that y = 5k-1. Using this and substituting into the series we get,$$\sum_{1}^{\infty}kr^{5k-1} =\sum_{4}^{\infty}\frac{y+1}{5} r^y$$The thing to notice here is the term on the right hand side it is derivative of r^(y+1). After we have the value of r (as a positive rational zero of P(x) ) we can simply substitute it into the final expression we get from the sum of the modified series. These are steps 2 and 3 solved through this method, step I is finding the root, which is still incomplete.

anonymous · Answer

Well I get the basic premise now. One of my friends told me that he did it without finding the root first. Which got me curious. Thanks for the help guys.