EX1: ak= c(n,k), k=0,1,2,,3,…,n
By definition, c(n,k)
=n!/(k!(n-k)!) and then c(5,2)=5!/(2!3!)=10
G(x)= S(0<=k<=n)
c(n,k) xk =
(x+1)n
EX2: ak = 1 for k=0,1,2,3,…, infinity
G(x) = 1 + x + x2 + x3 + x4 + …
= 1/(1-x) provided that abs(x)<1
EX3: ak = 1/k! for k=0,1,2,3,…, infinity
G(x) = 1/0! + x/1! + x2/2! + x3/3! + x4/4! + … = ex
Above equality is based upon the Taylor’s series of f(x) about x=0 which says:
f(x)
= f(0) + f ’(0) x/1! + f ’’(0) x2/2! + f ’’’(0) x3/3!
+ f ’’’’(0) x4 /4! + …+ f (k)(0) xk/k! + …