Chapter 9, Problem 66RE

To determine

To proof: The expression $1+\frac{3}{2}+2+\frac{5}{2}+\cdots +\frac{1}{2}\left(n+1\right)=\frac{n}{4}\left(n+3\right)$ with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

The expression $1+\frac{3}{2}+2+\frac{5}{2}+\cdots +\frac{1}{2}\left(n+1\right)=\frac{n}{4}\left(n+3\right)$ .

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 $1+\frac{3}{2}+2+\frac{5}{2}+\cdots +\frac{1}{2}\left(n+1\right)=\frac{n}{4}\left(n+3\right)$ .

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,

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

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 $n=k$ .

  $1+\frac{3}{2}+2+\frac{5}{2}+\cdots +\frac{1}{2}\left(k+1\right)=\frac{k}{4}\left(k+3\right)$

  ...(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 $S_k$ .

  $1+\frac{3}{2}+2+\frac{5}{2}+\cdots +\frac{1}{2}\left(k+1\right)+\frac{1}{2}\left[\left(k+1\right)+1\right]=S_k+\frac{1}{2}\left[\left(k+1\right)+1\right]$

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

  $\begin{array}{c}1+\frac{3}{2}+2+\frac{5}{2}+\cdots +\frac{1}{2}\left(k+1\right)+\frac{1}{2}\left[\left(k+1\right)+1\right]=S_k+\frac{1}{2}\left[\left(k+1\right)+1\right]\\ =\frac{k}{4}\left(k+3\right)+\frac{1}{2}\left[\left(k+1\right)+1\right]\\ =\frac{k^2+3k+2k+4}{4}\\ =\frac{k^2+5k+4}{4}\end{array}$

Factor out the terms and simplify,

  $\begin{array}{c}1+\frac{3}{2}+2+\frac{5}{2}+\cdots +\frac{1}{2}\left(k+1\right)+\frac{1}{2}\left[\left(k+1\right)+1\right]=\frac{k^2+4k+k+4}{4}\\ =\frac{k\left(k+4\right)+1\left(k+4\right)}{4}\\ =\frac{\left(k+1\right)\left(k+4\right)}{4}\\ =\frac{\left(k+1\right)\left[\left(k+1\right)+3\right]}{4}\end{array}$

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

  $1+\frac{3}{2}+2+\frac{5}{2}+\cdots +\frac{1}{2}\left(k+1\right)+\frac{1}{2}\left[\left(k+1\right)+1\right]=\frac{1}{4}\left(k+1\right)\left[\left(k+1\right)+3\right]$

Hence, by the principle of mathematical induction the statement $1+\frac{3}{2}+2+\frac{5}{2}+\cdots +\frac{1}{2}\left(n+1\right)=\frac{n}{4}\left(n+3\right)$ is true for all integers $n\ge 1$ .