Chapter 11.4, Problem 4E

To determine

To prove:

Given statement is true using principle of mathematical induction.

Explanation of Solution

Give information:

  $1.2+2.3+3.4+\mathrm{...}+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}$

Formula Used:

Principle of mathematical Induction.

Proof:

It is given that,

  $1.2+2.3+3.4+\mathrm{...}+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}$

Let $P\left(n\right)$ be $1.2+2.3+3.4+\mathrm{...}+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}$

By using principle of mathematical induction,

Put $n=1$ ,

  $1.2+2.3+3.4+\mathrm{...}+n\left(n+1\right)=\frac{1\left(1+1\right)\left(1+2\right)}{3}=\frac{1\left(2\right)\left(3\right)}{3}=\frac{6}{3}=2$

  $P\left(1\right)$ is true , as both sides of the equations is equal to $2$ .

Let $P\left(k\right)$ be true.

  $1.2+2.3+3.4+\mathrm{...}+k\left(k+1\right)=\frac{k\left(k+1\right)\left(k+2\right)}{3}$

Need to prove that $P\left(k+1\right)$ is also true.

  $\begin{array}{l}1.2+2.3+3.4+\mathrm{...}+k\left(k+1\right)\left(k+1\right)\left(k+2\right)\\ =\frac{k\left(k+1\right)\left(k+2\right)}{3}+\left(k+1\right)\left(k+2\right)\\ =\left(\frac{k}{3}+1\right)\left(k+1\right)\left(k+2\right)\\ =\left(\frac{k}{3}+\frac{3}{3}\right)\left(k+1\right)\left(k+2\right)\\ =\left(\frac{k+3}{3}\right)\left(k+1\right)\left(k+2\right)\\ =\left(\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\right)\\ =\left(\frac{\left(k+1\right)\left(\left(k+1\right)+1\right)\left(k+1\right)+2}{3}\right)\end{array}$

Therefore, $P\left(k+1\right)$ is also true.

Therefore, by principle of mathematical induction $P\left(n\right)$ is true for all positive integers $n$

Given statement is true.