# Suppose that $a_n = a_0r^n$. This is a geometric sequence, since $a_{n+1}/a_n = r$ is independent of n.

Suppose that $a_n = a_0r^n$. This is a geometric sequence, since $a_{n+1}/a_n = r$ is independent of n.

a) Give the generating function $A(X) = \sum_{n}a_nX^n$
b) Give the generating function $B(X) = \sum_{n}b_nX^n$ with $b_n = \sum_{k=0}^n a_k$
c) Calculate $[X^n]B(X)$, and thus give a formula for the sum of a finite geometric sequence.

For Part a) what I thought to do from my notes was
$A(X) + A(−X) = (a_0 + a_1X + a_2X^2 + a_3X^3 +...) +(a_0 + a_1(−X) + a_2(−X)^2 + a_3(−X)^3 +...)$ $= (a_0 + a_1X + a_2X^2 + a_3X^3 +...) + (a_0 − a_1X + a_2X^2 − a_3X^3 +...)$
$= (2a_0 + a_2X^2 + a_4X^4 +...)$

Which gives the generating function for the even terms and you can use it to get the odd terms with a small change. Im not sure if this is right however.

• Is $a_n$ defined for $n = 0,1,2,3,\ldots$, or for $n = \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots$, or something else? – Bungo Nov 30 '16 at 1:12
• Im not sure but from what i've seen in class it has always been n>= 0 – KGT Nov 30 '16 at 1:28
• I'm not sure what you are doing for part (a). Isn't the generating function associated with a sequence $(a_n)_{n=0}^{\infty}$ simply the series $\sum_{n=0}^{\infty}a_n X^n$? In your case this will be $\displaystyle a_0 \sum_{n=0}^{\infty}r^n X^n = a_0\sum_{n=0}^{\infty}(rX)^n = \frac{a_0}{1-rX}$, where the last equality holds provided that $|rX| < 1$, or equivalently, $|X| < 1/|r|$ (assuming $r$ is nonzero). – Bungo Nov 30 '16 at 1:31