Chapter 11.7, Problem 31PPS

To determine

Prove that $\frac{1}{1·2·3}+\frac{1}{2·3·4}+\frac{1}{3·4·5}+\mathrm{....}+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ is true for all natural numbers.

Explanation of Solution

Given:

Prove that $\frac{1}{1·2·3}+\frac{1}{2·3·4}+\frac{1}{3·4·5}+\mathrm{....}+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ is true for all natural numbers.

Concept Used:

Prove by mathematical induction.

Calculation:

Step 1: When n = 1, the left side of the equation is $\frac{1}{1(1+1)(1+2)}=\frac{1}{6}$ .

The right side is $\frac{1(1+3)}{4(1+1)(1+2)}=\frac{1}{6}$ , so the equation is true for n = 1

Step 2: Assume that

  $\frac{1}{1·2·3}+\frac{1}{2·3·4}+\frac{1}{3·4·5}+\mathrm{....}+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)}$ for some natural number k.

Step 3:

Add: $\frac{1}{(k+1)(k+2)(k+3)}$

  $\begin{array}{l}\frac{1}{1·2·3}+\frac{1}{2·3·4}+\frac{1}{3·4·5}+\mathrm{....}+\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}\\ =\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}\\ =\frac{k(k+3)(k+3)}{4(k+1)(k+2)(k+3)}+\frac{1}{(k+1)(k+2)(k+3)}\\ =\frac{k^3+6k^2+9k+4}{4(k+1)(k+2)(k+3)}\\ =\frac{(k+1)(k+1)(k+4)}{4(k+1)(k+2)(k+3)}=\frac{(k+1)[(k+1)+3]}{4[(k+1)+1][(k+1)+2]}\end{array}$

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

  $\frac{1}{1·2·3}+\frac{1}{2·3·4}+\frac{1}{3·4·5}+\mathrm{....}+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ for all natural number n .