in Combinatory
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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.

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last 3rd line :P
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1 Answer

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Best answer

(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

edited by

2 Comments

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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edited by
thanks bro...
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