Chapter 11, Problem 63SGR

To determine

Prove that the statement $2+6+12+\mathrm{.....}+n(n+1)=\frac{n(n+1)(n+2)}{3}$ is true for all positive integers.

Answer to Problem 63SGR

Explanation of Solution

Given:

The expression: $2+6+12+\mathrm{.....}+n(n+1)=\frac{n(n+1)(n+2)}{3}$

Concept Used:

A proof by induction consists of two cases. The first, the base case (or basis), proves the statement for n = 1 without assuming any knowledge of other cases. The second case, the induction step, proves that if the statement holds for any given case n = k, then it must also hold for the next case n = k + 1.

Calculation:

Step 1: When n = 1 , the left side of the given equation is 2.

The right side is $\frac{1(1+1)(1+2)}{3}=2$ , so the equation is true for n = 1

Step 2: Assume that $2+4+6+\mathrm{.....}+k(k+1)=\frac{k(k+1)(k+2)}{3}$ for some natural number k.

Step 3: $2+4+6+\mathrm{.....}+k(k+1)+(k+1)(k+2)$

  $\begin{array}{l}=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)\\ =(k+1)(k+2)[\frac{k}{3}+1]=\frac{(k+1)(k+2)(k+3)}{3}\end{array}$

The last expression is the right side of the equation to be proved, where $n=k+1$ .

Therefore, $2+4+6+\mathrm{.....}+k(k+1)=\frac{k(k+1)(k+2)}{3}$ for all natural numbers n.

Thus, $2+4+6+\mathrm{.....}+k(k+1)=\frac{k(k+1)(k+2)}{3}$ for all natural numbers n.