I got (x^20).1/(1+x).1/(1-x)^3 . Now, if I write this as - [x^20].(1+x)^-1.(1-x)^-3
Now, I don't know how to solve this further when we have both factors having negative n.
Please don't send me here - https://gateoverflow.in/65803/find-number-integral-solutions-using-generating-function
As i Didn't get that solution.
Someone please simplify it.
(1+x)^-1 = 1 - x + x^2 - x^3 + x^4.... =$\sum_{r=0}^{r=\infty }-1^{r}*x^{r}$
(Refer http://mathworld.wolfram.com/NegativeBinomialSeries.html )
(1-x)^-3 = = 1 + 3x + 6x^2 + 10x^3 + 15x^4 + . = (n + 1)(n + 2)/ 2 * (x^n)
(Refer : https://www.gotohaggstrom.com/The%20binomial%20series%20for%20negative%20integral%20exponents.pdf )
Rest is same as https://gateoverflow.in/65803/find-number-integral-solutions-using-generating-function by applying binomial theorm
also for generating function see https://www.youtube.com/watch?v=4d2XEn1j_q4&list=PL0862D1A947252D20&index=30
https://gateoverflow.in/?qa=blob&qa_blobid=8520527711789931294
here you have done some mistakes
(1+x)^-3 = = 1 - 3x + 6x^2 - 10x^3 + 15x^4 + .....
(1-x)^-3 = = 1 + 3x + 6x^2 + 10x^3 + 15x^4 + . ......
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