Find the sum of the infinite geometric series. $a + a x ^ { 2 } + a x ^ { 4 } + a x ^ { 6 } + \ldots$
A) $S = \frac { x } { 1 - \alpha ^ { 2 } }$

B) $S = \frac { \alpha } { 1 - x ^ { 4 } }$

C) $S = \frac { x ^ { 2 } } { 1 - a ^ { 2 } }$

D) $S = \frac { 1 } { 1 + x ^ { 2 } }$

E) $S = \frac { \alpha } { 1 - x ^ { 2 } }$

Question

Find the sum of the infinite geometric series. $a + a x ^ { 2 } + a x ^ { 4 } + a x ^ { 6 } + \ldots$
A) $S = \frac { x } { 1 - \alpha ^ { 2 } }$

B) $S = \frac { \alpha } { 1 - x ^ { 4 } }$

C) $S = \frac { x ^ { 2 } } { 1 - a ^ { 2 } }$

D) $S = \frac { 1 } { 1 + x ^ { 2 } }$

E) $S = \frac { \alpha } { 1 - x ^ { 2 } }$

Quizplus · Accepted Answer

The sum of an infinite geometric series with first term $a$ and common ratio $r$ is $\frac{a}{1-r}$ if $|r|<1$. In this case, the first term is $a$, and the common ratio is $x^2$. Thus, the sum is $\frac{a}{1-x^2}$. However, none of the answer choices have this exact form. Therefore, substitute $\alpha$ for $a$ and simplify as follows: $$\frac{\alpha}{1-x^2}=\frac{\alpha}{(1+x)(1-x)}=\frac{\alpha}{1+x}\cdot\frac{1}{1-x}.$$ Thus, the correct answer choice is $\boxed{	extbf{(E)}}$.

Find the Sum of the Infinite Geometric Series a+ax2+ax4+ax6+…a + a x ^ { 2 } + a x ^ { 4 } + a x ^ { 6 } + \ldotsa+ax2+ax4+ax6+…

Find the Sum of the Infinite Geometric Series $a + a x ^ { 2 } + a x ^ { 4 } + a x ^ { 6 } + \ldots$