Chapter 9.4, Problem 15E

To determine

To proof: The expression $1+2+2^2+2^3+\cdots +2^{n-1}=2^n-1$ with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

The expression $1+2+2^2+2^3+\cdots +2^{n-1}=2^n-1$ .

Formula used:

The steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Proof:

Consider expression $1+2+2^2+2^3+\cdots +2^{n-1}=2^n-1$ .

Denote the expression by $S_n$ .

Recall that the steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Substitute $n=1$ in the expression,

  $\begin{array}{c}{2}^{1-1}=2^1-1\\ 2^0=2-1\\ 1=1\end{array}$

The above equation holds true.

Therefore, the statement is true for $n=1$ so $S_1$ is true.

Now, assume that the statement if true for $n=k$ .

  $1+2+2^2+2^3+\cdots +2^{k-1}=2^k-1$  ...(1)

Therefore, the statement holds true for $n=k$ so $S_k$ is true.

Now, prove it for $n=k+1$ . Write the expression for $n=k+1$ . Group the terms that involve $S_k$ .

  $1+2+2^2+2^3+\cdots +2^{k-1}+2^k=S_k+2^k$

Substitute the value of $S_k$ from equation (1),

  $\begin{array}{c}1+2+2^2+2^3+\cdots +2^{k-1}+2^k=S_k+2^k\\ =2^k-1+2^k\\ =2\cdot 2^k-1\\ =2^{k+1}-1\end{array}$

Result obtained is that $S_k$ implies $S_{k+1}$ that is,

  $1+2+2^2+2^3+\cdots +2^{k-1}+2^k=2^{k+1}-1$

Hence, by the principle of mathematical induction the statement $1+2+2^2+2^3+\cdots +2^{n-1}=2^n-1$ is true for all integers $n\ge 1$ .