Chapter 11.7, Problem 11PPS

To determine

To prove that for each natural number

  $1+5+9+\mathrm{....}+\left(4n-3\right)=n\left(2n-1\right)$

Explanation of Solution

Given:

  $1+5+9+\mathrm{....}+\left(4n-3\right)=n\left(2n-1\right)$

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

  $1+2+4+\mathrm{....}+2^{k-1}+2^{\left(k+1\right)-1}=2^{k+1}-1$

Use mathematical induction.

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

For, the left hand side of given identity is:

  $\left(4\times 1-3\right)=1$

And right hand side is:

  $\begin{array}{c}n\left(2n-1\right)=1\left(2\times 1-1\right)\left(Puttingn=1\right)\\ =1\\ =\text{Left side}\end{array}$

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

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

  $1+5+9+\mathrm{....}+\left(4k-3\right)=k\left(2k-1\right)$

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

  $1+5+9+\mathrm{....}+\left(4k-3\right)+\left(4\left(k+1\right)-3\right)=\left(k+1\right)\left(2\left(k+1\right)-1\right)$

To prove above simplify the left side of the above, as

  $\begin{array}{c}1+5+9+\mathrm{....}+\left(4k-3\right)+\left(4\left(k+1\right)-3\right)=k\left(2k-1\right)+\left(4k+4-3\right)\because 1+5+9+\mathrm{....}+\left(4k-3\right)=k\left(2k-1\right)\\ =2k^2-k+4k+1\\ =2k^2+3k+1\\ =2k^2+2k+k+1\\ =2k\left(k+1\right)+1\left(k+1\right)\\ =\left(k+1\right)\left(2k+1\right)\\ =\left(k+1\right)\left(2\left(k+1\right)-1\right)\\ =\text{Right side}\end{array}$

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

  $1+5+9+\mathrm{....}+\left(4n-3\right)=n\left(2n-1\right)$