Solve the problem.
-Find the sum of the infinite series $1 + 6 r + r ^ { 2 } + 6 r ^ { 3 } + r ^ { 4 } + 6 r ^ { 5 } + r ^ { 6 } + \ldots$ for those values of $r$ for which it converges.
A) $\frac { 1 + 6 r } { 1 + r ^ { 2 } }$
B) $\frac { 1 - 6 r } { 1 + r ^ { 2 } }$
C) $\frac { 1 - 6 r } { 1 - r ^ { 2 } }$
D) $\frac { 1 + 6 r } { 1 - r ^ { 2 } }$

Question

Quizplus · Accepted Answer

The given series alternates between terms of the form $1, r^2, r^4, \ldots$ and $6r, 6r^3, 6r^5, \ldots$. These can be seen as two separate geometric series: $1 + r^2 + r^4 + \ldots$ and $6r + 6r^3 + 6r^5 + \ldots$. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$, where $a$ is the first term and $r$ is the common ratio. For the first series, $a = 1$ and $r = r^2$, giving a sum of $\frac{1}{1 - r^2}$. For the second series, $a = 6r$ and $r = r^2$, giving a sum of $\frac{6r}{1 - r^2}$. Adding these sums gives the total sum of the series as $\frac{1}{1 - r^2} + \frac{6r}{1 - r^2} = \frac{1 + 6r}{1 - r^2}$, which matches option D.

Solve the Problem 1+6r+r2+6r3+r4+6r5+r6+…1 + 6 r + r ^ { 2 } + 6 r ^ { 3 } + r ^ { 4 } + 6 r ^ { 5 } + r ^ { 6 } + \ldots1+6r+r2+6r3+r4+6r5+r6+…

Solve the Problem $1 + 6 r + r ^ { 2 } + 6 r ^ { 3 } + r ^ { 4 } + 6 r ^ { 5 } + r ^ { 6 } + \ldots$