AP Calculus BC Geometric Series Problem With Solution

AP Calculus BC Geometric Series Problem

With Solution

Yesterday I gave an AP Calculus BC problems involving a geometric series. You can find that post here: AP Calculus BC Problem – Geometric Series.

Today I will provide a solution to that problem. First, here is the question one more time.

Level 2 Series

$g(x)=\Sigma_{n=1}^{\infty}(sin^2 x)^n$

If g is the function given above, then g(π/3) =

(A) 3
(B) 1
(C) 3/4
(D) 1/2

Solution:
$g(\frac{\pi}{3})$
$=\Sigma_{n=1}^{\infty}(sin^2(\frac{\pi}{3}))^n$
$=\Sigma_{n=1}^{\infty}((\frac{\sqrt{3}}{2})^2)^n$
$=\Sigma_{n=1}^{\infty}(\frac{3}{4})^n$
$=\frac{\frac{3}{4}}{1-\frac{3}{4}}$
$=\frac{3}{4}\div \frac{1}{4}$
$=\frac{3}{4}\cdot 4$
$=3$
Note:
$\Sigma_{n=1}^{\infty}(\frac{3}{4})^n$