Printable View
The question is as follows:
It is known that the generating function for the sequence , {1,8,27,64,...} is
[x(x^2 + 4x +1)]/(1-x)^4
Use the result to show summation from r=1 to n of r^3 = [n^2(n^2+1)]/4
I know i have to find the sequence represented by 1,8,27,64, .. and once i get the sequence i have to manipulate it until i reach r^3 which will be given by the expression.
My problem is that i cannot find the sequence for 1,8,27,64,...
any help please???
Thank buches!
Hello, mehn!
Please check the wording of the problem.
None of it makes sense.
Quote:
It is known that the generating function for the sequence , {1, 8, 27, 64, ...}
. . is: .[x(x^2 + 4x +1)]/(1-x)^4 . What?
Use the result to show summation from r = 1 to n of r^3
. .is: .[n^2(n^2+1)]/4 . Not true
The generating function is: .f(r) = r^3
. . The terms are consecutive cubes.
The sum of the first n cubes is: .[n^2(n+1)^2]/4
Please read Generating Functions. The OP's generating function for cubes is correct.Quote:
Originally Posted by Soroban
This is a strange problem (to me, at any rate), but it actually works!Quote:
Originally Posted by mehn
The binomial series for (1-x)^{-4} is 1 + 4x + 10x^2 + 20x^3 + ... . When you multiply this by x(x^2 + 4x +1), you get x + 8x^2 + 27x^3 + 64x^4 + ... . The coefficients magically appear as the cubes of the integers!
In this problem, you are asked to find the sum of the first n coefficients. One way to do that is to multiply the given generating function by (1-x)^{-1} and looking for the coefficient of x^n in that product.
The coefficient of x^n in (1-x)^{-5} is http://latex.codecogs.com/png.latex?\frac{(n+4)!}{4!n!}. So the coefficient of x^n in http://latex.codecogs.com/png.latex?... +1)}{(1-x)^5} is
http://latex.codecogs.com/png.latex?...n-1)!}\right),
which eventually simplifies to [n^2(n+1)^2]/4.
Not the easiest way to get that formula, but an interesting piece of mathematics.
Edit. Thanks to alexmahone for that useful link. I hadn't seen his comment when writing mine.