2 votes 2 votes Let $G(x)= \frac{1}{(1-x)} =\sum_{0}^{\infty}f(i)x^i$ where $|x|< 1$. $f(i)$ is $1$ $i$ $2i$$i+1$ Combinatory go2025-dm-2 generating-functions combinatory + – gatecse asked Jun 28, 2020 gatecse 196 views answer comment Share Follow See 1 comment See all 1 1 comment reply theredeepakb commented Dec 18, 2021 reply Follow Share G(x)is a function of series which are in G.P 0 votes 0 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes The expansion of $\frac{1}{1-x}$ is $1+x+x^2+x^3+\ldots$ So, $f(i)$ is the constant function $1.$ Correct answer is A. gatecse answered Jun 28, 2020 • selected Jul 23, 2020 by Lakshman Bhaiya gatecse comment Share Follow See all 0 reply Please log in or register to add a comment.