Chapter 11.7, Problem 1CYU

To determine

To prove

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

Explanation of Solution

Given:

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

Concept Used:

An identity can be proved using mathematical induction in two steps:

1. First prove the identity is true for $n=1$ .
2. Prove that if it is true for $n=k$ then it is also true for $n=k+1$ .

Calculation:

In order to prove

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

Use mathematical induction.

First, let’s proof that the identity is true for $n=1$ .

For, the left hand side of given identity is:

  $1^{}=1$

And right hand side is:

  $\begin{array}{c}{n}^2\left(\text{Putting }n=1\right)\\ =1^2\\ =1\\ =\text{Left side}\end{array}$

Thus, the given identity is true for $n=1$ .

Now, let it is true for $n=k;\text{ for some }k>1$ . Thus,

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

Now, let’s show that it is also true for $n=k+1$ , i.e. prove

Adding $2k+1$ on both sides

Take left side of above and simplify as

  $\begin{array}{c}1+3+5+\mathrm{......}+\left(2k-1\right)+\left(2k+2-1\right)={\left(k+1\right)}^2\\ 1+3+5+\mathrm{......}+\left(2k-1\right)+\left(2\left(k+1\right)-1\right)={\left(k+1\right)}^2\end{array}$

Thus, the identity is also true for $n=k+1$ , when it is true for $n=k$ .

Thus by mathematical induction it is proved that

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