Question

In Problem, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers $n$.

role="math" localid="1647675344162" $1·2+2·3+3·4+...+n(n+1)=\frac{1}{3}n(n+1)(n+2)$

Short answer

The given statement has been shown.

Step-by-step solution

Step 1. Given information

Statement is$1·2+2·3+3·4+...+n(n+1)=\frac{1}{3}n(n+1)(n+2)$

Step 2. Show the given statement using the principle of mathematical induction.

For $n=1$

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

Assume that the statement is true for $n=k$.

$1·2+2·3+3·4+\dots \dots \dots \dots +k(k+1)=\frac{1}{3}k(k+1)(k+2)$

style="max-width: none; vertical-align: -89px;" width="552" $\begin{array}{rcl}\frac{1}{3}(k+1)\left(\right(k+1)+1)\left(\right(k+1)+2)& =& 1·2+2·3+\dots \dots \dots +k(k+1)+(k+1)\left(\right(k+1)+1)\\ & =& \frac{1}{3}k(k+1)(k+2)+(k+1)(k+2)\\ & =& \frac{1}{3}(k+1)(k+1+1)[k+1+2]\\ & =& \frac{1}{3}(k+1)\left[\right(k+1)+1]\left[\right(k+1)+2]\end{array}$

Thus, the statement is true for k + 1.

Since both the conditions of the Principle of Mathematical Induction are satisfied, the statement is true for all natural numbers.