3 votes 3 votes 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. Combinatory combinatory generating-functions discrete-mathematics + – iarnav asked Jul 16, 2017 iarnav 2.0k views answer comment Share Follow See 1 comment See all 1 1 comment reply sumit goyal 1 commented Jan 28, 2018 reply Follow Share last 3rd line :P 0 votes 0 votes Please log in or register to add a comment.
Best answer 3 votes 3 votes (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 Aashish S answered Jul 16, 2017 • edited Jan 28, 2018 by Aashish S Aashish S comment Share Follow See all 2 Comments See all 2 2 Comments reply Prateek kumar commented Jan 28, 2018 reply Follow Share 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 + . ...... 0 votes 0 votes Aashish S commented Jan 28, 2018 i edited by Aashish S Jan 28, 2018 reply Follow Share thanks bro... 0 votes 0 votes Please log in or register to add a comment.