Chapter 9, Problem 72RE

Solution

To prove: The given statement is true or not for any positive integer using mathematical induction.

Given information:

The given statement is $1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}$ .

Formula used :

Follow the steps to prove that the statement is true for any positive integer.

Anchor step: Prove that any statement $P_n$ is true for $n=1$ .

Inductive hypothesis: Assume $P_n$ is true for $n=k$ .

Inductive Step: Prove that $P_n$ is true for $n=k+1$ .

Explanation:

Conjecture:

Take $P_n$ to be statement $1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}$ .

Anchor step:

Prove that $P_n$ is true for $n=1$ .

For $n=1$ ,

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

Inductive hypothesis:

Assume $P_n$ is true for $n=k$ . Replace $n$ with $k$ in $P_n$ .

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

Inductive Step:

Prove that $P_n$ is true for $n=k+1$ .

This means it is needed to be proved that $P_{k+1}$ must be true.

Proceed with the inductive hypothesis.

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

Add ${\left(k+1\right)}^{th}$ term on both sides and simplify the right-hand side.

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

It can be written as:

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

The statement is same as the statement $P_{k+1}$ , so $P_{k+1}$ is true.

So, if $P_n$ is true for $n=k$ , then it is also true for $n=k+1$ .

Conclusion:

As $P_n$ is true for $n=1$ and $P_k$ further implies $P_{k+1}$ , this means $P_n$ is true for $n=2$ , $n=3$ and so on.

By mathematical induction, $1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}$ is true for all positive integers $n$ .