🚀 go-pugleaf

>I'm wondering if there is an expression for F(n) = a**0 + a**1 + ... + a**n
>in term of both a and n. Here a is a constant. I know the expression for
>a = 2, which is 2**(n+1) -1.  But now I want to know the expression for
>any constant a.  Thanks for any help.

There is a simple trick for finding a closed form expression for your
function F(n).  First, if a=1, it is clear that F(n) = n+1.  Now
consider the case where a is not 1.  Then a-1 is nonzero, and we can
multiply and divide F(n) by a-1.  Expanding and simplifying the
numerator gives the desired result (all but two of the terms in the
numerator will cancel):

F(n) = (a - 1) * (a**0 + a**1 + ... + a**n) / (a - 1)
     = {(a**1 + a**2 + ... + a**(n+1)) - (a**0 + a**1 + ... + a**n)} / (a - 1)
     = (a**(n+1) - 1) / (a - 1)

Your function F(n) is referred to as "the sum of a finite geometric
series", or as "the nth partial sum of an infinite geometric series".


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