Chapter 9, Problem 10CT

To determine

To proof: The expression $5+10+15+\cdots +5n=\frac{5n\left(n+1\right)}{2}$ with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

The expression $5+10+15+\cdots +5n=\frac{5n\left(n+1\right)}{2}$ .

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 $5+10+15+\cdots +5n=\frac{5n\left(n+1\right)}{2}$ .

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,

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

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 $n=k$ .

  $5+10+15+\cdots +5k=\frac{5k\left(k+1\right)}{2}$ …(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 $S_k$ .

  $5+10+15+\cdots +5k+5\left(k+1\right)=S_k+5\left(k+1\right)$

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

  $\begin{array}{c}5+10+15+\cdots +5k+5\left(k+1\right)=S_k+5\left(k+1\right)\\ =\frac{5k\left(k+1\right)}{2}+5\left(k+1\right)\\ =5\left(k+1\right)\left(\frac{k}{2}+1\right)\end{array}$

Simplify it further as,

  $\begin{array}{c}5+10+15+\cdots +5k+5\left(k+1\right)=5\left(k+1\right)\left(\frac{k}{2}+1\right)\\ =\frac{5\left(k+1\right)\left(k+2\right)}{2}\\ =\frac{5\left(k+1\right)\left[\left(k+1\right)+1\right]}{2}\end{array}$

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

  $5+10+15+\cdots +5k+5\left(k+1\right)=\frac{5\left(k+1\right)\left[\left(k+1\right)+1\right]}{2}$

Hence, by the principle of mathematical induction the statement $5+10+15+\cdots +5n=\frac{5n\left(n+1\right)}{2}$ is true for all integers $n\ge 1$ .