Chapter 9.4, Problem 19E

To determine

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

Explanation of Solution

Given information:

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

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

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,

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

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 $n=k$ .

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

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

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

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

Factor out the terms and simplify,

  $\begin{array}{c}{1}^2+3^2+5^2+\cdots +{\left(2k-1\right)}^2+{\left[2\left(k+1\right)-1\right]}^2=\left(2k+1\right)\left[\frac{k\left(2k-1\right)}{3}+2k+1\right]\\ =\left(2k+1\right)\left(\frac{2k^2-k+6k+3}{3}\right)\\ =\left(2k+1\right)\left(\frac{2k^2+5k+3}{3}\right)\\ =\left(2k+1\right)\left(\frac{2k^2+2k+3k+3}{3}\right)\end{array}$

Rewrite the above expression as,

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

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

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

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