Question

b2 - series: problem 9
(1 point)
what is the value of c if \\(\sum_{n = 1}^{\infty}(1 + c)^{-n}=3\\) and \\(c>0\\)?
answer: c =

Explanation:

Step1: Recognize the geometric - series formula

The sum of an infinite geometric series $\sum_{n = 1}^{\infty}ar^{n - 1}=\frac{a}{1 - r}$ when $|r|\lt1$. Rewrite $\sum_{n = 1}^{\infty}(1 + c)^{-n}$ as $\sum_{n = 1}^{\infty}\frac{1}{(1 + c)^{n}}$. Let $r=\frac{1}{1 + c}$ and $a=\frac{1}{1 + c}$, then $\sum_{n = 1}^{\infty}(1 + c)^{-n}=\sum_{n = 1}^{\infty}\frac{1}{1 + c}\cdot(\frac{1}{1 + c})^{n - 1}$.

Step2: Apply the geometric - series sum formula

Using the sum formula for an infinite geometric series $S=\frac{a}{1 - r}$, with $a=\frac{1}{1 + c}$ and $r=\frac{1}{1 + c}$, we have $\sum_{n = 1}^{\infty}(1 + c)^{-n}=\frac{\frac{1}{1 + c}}{1-\frac{1}{1 + c}}$.

Step3: Simplify the expression

First, simplify the denominator: $1-\frac{1}{1 + c}=\frac{1 + c-1}{1 + c}=\frac{c}{1 + c}$. Then $\frac{\frac{1}{1 + c}}{1-\frac{1}{1 + c}}=\frac{\frac{1}{1 + c}}{\frac{c}{1 + c}}=\frac{1}{c}$.

Step4: Solve for $c$

Since $\sum_{n = 1}^{\infty}(1 + c)^{-n}=3$, and we found that $\sum_{n = 1}^{\infty}(1 + c)^{-n}=\frac{1}{c}$, we set up the equation $\frac{1}{c}=3$. Cross - multiply to get $3c = 1$, so $c=\frac{1}{3}$.

Answer:

$c=\frac{1}{3}$