Chapter 9.4, Problem 13E

To determine

To proof: The expression $2+7+12+17+\cdots +\left(5n-3\right)=\frac{n}{2}\left(5n-1\right)$ with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

The expression $2+7+12+17+\cdots +\left(5n-3\right)=\frac{n}{2}\left(5n-1\right)$ .

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 $2+7+12+17+\cdots +\left(5n-3\right)=\frac{n}{2}\left(5n-1\right)$ .

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,

  $\begin{array}{c}5\left(1\right)-3=\frac{1}{2}\left(5\left(1\right)-1\right)\\ 5-3=\frac{4}{2}\\ 2=2\end{array}$

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 $n=k$ .

  $2+7+12+17+\cdots +\left(5k-3\right)=\frac{k}{2}\left(5k-1\right)$  ...(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 $S_k$ .

  $2+7+12+17+\cdots +\left(5k-3\right)+\left(5\left(k+1\right)-3\right)=S_k+\left(5\left(k+1\right)-3\right)$

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

  $\begin{array}{c}2+7+12+17+\cdots +\left(5k-3\right)+\left(5\left(k+1\right)-3\right)=S_k+\left(5\left(k+1\right)-3\right)\\ =\frac{k}{2}\left(5k-1\right)+\left(5\left(k+1\right)-3\right)\end{array}$

Simplify the terms obtained,

  $\begin{array}{c}2+7+12+17+\cdots +\left(5k-3\right)+\left(5\left(k+1\right)-3\right)=\frac{k}{2}\left(5k-1\right)+5k+2\\ =\frac{5k^2-k+10k+4}{2}\\ =\frac{5k^2+9k+4}{2}\\ =\frac{5k^2+5k+4k+4}{2}\end{array}$

Factor out the common terms,

  $\begin{array}{c}2+7+12+17+\cdots +\left(5k-3\right)+\left(5\left(k+1\right)-3\right)=\frac{5k\left(k+1\right)+4\left(k+1\right)}{2}\\ =\frac{\left(k+1\right)\left(5k+4\right)}{2}\\ =\frac{\left(k+1\right)\left(5k+5-1\right)}{2}\\ =\frac{\left(k+1\right)\left(5\left(k+1\right)-1\right)}{2}\end{array}$

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

  $2+7+12+17+\cdots +\left(5k-3\right)+\left(5\left(k+1\right)-3\right)=\frac{\left(k+1\right)\left(5\left(k+1\right)-1\right)}{2}$

Hence, by the principle of mathematical induction the statement $2+7+12+17+\cdots +\left(5n-3\right)=\frac{n}{2}\left(5n-1\right)$ is true for all integers $n\ge 1$ .