Chapter 12.9, Problem 13E

To determine

To prove: that the given proposition is valid for all positive integral values of $n$ by using mathematical induction.

Answer to Problem 13E

See explanation

Explanation of Solution

Given information:

  $-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\mathrm{...}-\frac{1}{2^n}=\frac{1}{2^n}-1$

Calculation:

Let the given proposition be $S_n$ .

  $S_n=\frac{1}{2^n}-1$

Now verify that $S_n$ is valid for $n=1$ .

  $\begin{array}{c}{S}_1=\frac{1}{2^1}-1\\ =-\frac{1}{2}\end{array}$

Since $S_1$ is the first term of the proposition. So $S_n$ is valid for $n=1$ .

Assume that $S_n$ is valid for $n=k$ , i.e. $S_k=-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\mathrm{...}-\frac{1}{2^k}=\frac{1}{2^k}-1$ .

Now verify that $S_n$ is valid for $n=k+1$ as shown below.

  $\begin{array}{c}{S}_{k+1}=S_k+\frac{1}{2^{k+1}}\\ =\frac{1}{2^k}-1+\frac{1}{2^{k+1}}\\ =\frac{1}{2^{k+1}}-1\end{array}$

Put the value of $n=k+1$ in $S_n$ as follows:

  $\begin{array}{c}{S}_n=\frac{1}{2^{k+1}}-1\\ =S_{k+1}\end{array}$

Thus, if the given proposition is valid for $n=k$ then it is also valid for $n=k+1$ . Since the given proposition is valid $n=1$ then it is also valid for $n=2,n=3$ and so on. Thus, the given statement is valid for all integers.