Question

(b) $sum_{n = 2}^{infty}\frac{1}{2^{n}}=$(c) $sum_{n=0}^{infty}\frac{2^{n}}{7^{2n + 1}}$

Explanation:

Step1: Identify the series type

The series \(\sum_{n = 2}^{\infty}\frac{1}{2^{n}}\) is a geometric - series with the general form of a geometric series \(\sum_{n = k}^{\infty}ar^{n - k}\), where \(a\) is the first - term and \(r\) is the common ratio. For \(\sum_{n = 2}^{\infty}\frac{1}{2^{n}}\), we can rewrite it as \(\sum_{n = 2}^{\infty}(\frac{1}{2})^{n}\). The common ratio \(r=\frac{1}{2}\). The formula for the sum of an infinite geometric series is \(S=\frac{a}{1 - r}\) when \(|r|\lt1\). First, find the first - term \(a\) when \(n = 2\), \(a=\frac{1}{2^{2}}=\frac{1}{4}\).

Step2: Calculate the sum of the geometric series

Using the sum formula \(S=\frac{a}{1 - r}\), with \(a=\frac{1}{4}\) and \(r=\frac{1}{2}\), we have \(S=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\).

For the series \(\sum_{n = 0}^{\infty}\frac{2^{n}}{7^{2n + 1}}\), rewrite it as \(\sum_{n = 0}^{\infty}\frac{2^{n}}{7\times49^{n}}=\frac{1}{7}\sum_{n = 0}^{\infty}(\frac{2}{49})^{n}\). Here, it is a geometric series with \(a = 1\) (when \(n = 0\)) and \(r=\frac{2}{49}\).

Step3: Calculate the sum of the second geometric series

Using the sum formula \(S=\frac{a}{1 - r}\) for the series \(\frac{1}{7}\sum_{n = 0}^{\infty}(\frac{2}{49})^{n}\), with \(a = 1\) and \(r=\frac{2}{49}\), we get \(S=\frac{1}{7}\times\frac{1}{1-\frac{2}{49}}=\frac{1}{7}\times\frac{49}{47}=\frac{7}{47}\).

Answer:

The sum of \(\sum_{n = 2}^{\infty}\frac{1}{2^{n}}\) is \(\frac{1}{2}\), and the sum of \(\sum_{n = 0}^{\infty}\frac{2^{n}}{7^{2n+1}}\) is \(\frac{7}{47}\).