Chapter 9.4, Problem 43E

To determine

To Find :The sum of the first n terms ,using mathematical induction

  $\frac{1}{4},\frac{1}{12},\frac{1}{24},\frac{1}{40},\mathrm{.......}\frac{1}{2n(n+1)},\mathrm{...}$

Answer to Problem 43E

The formula is $S_n=\frac{n}{2(n+1)}$

Explanation of Solution

Given: Use the mathematical induction to find the sum of the first n terms of the sequence .

  $\frac{1}{4},\frac{1}{12},\frac{1}{24},\frac{1}{40},\mathrm{.......}\frac{1}{2n(n+1)},\mathrm{...}$

The given , $\frac{1}{4},\frac{1}{12},\frac{1}{24},\frac{1}{40},\mathrm{.......}\frac{1}{2n(n+1)},\mathrm{...}$

Begin by writing first few sums

  $\begin{array}{l}{S}_n=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40},\mathrm{......}+\frac{1}{2n(n+1)}\\ s_1=\frac{1}{4}=\frac{1}{2(2)}\\ S_2=\frac{1}{4}+\frac{1}{12}=\frac{4}{12}=\frac{2}{6}=\frac{2}{2(3)}\\ S_3=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}=\frac{3}{8}=\frac{3}{2(4)}\\ S_4=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}=\frac{16}{40}=\frac{4}{10}=\frac{4}{2(5)}\end{array}$

It appears that The sum of n terms is

  $S_n=\frac{n}{2(n+1)}$

This can be verified by mathematical induction

The formula has already been verified for n=1

Assume that the formula is valid for n=k

Then

  $\begin{array}{l}{S}_{k+1}=\left[\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40},\mathrm{......}+\frac{1}{2k(k+1)}\right]+\frac{1}{2(k+1)(k+2)}\\ =\frac{k}{2(k+1)}+\frac{2}{2(k+1)(k+2)}\\ =\frac{k(k+2)+1}{2(k+1)(k+2)}\\ =\frac{k^2+2k+1}{2(k+1)(k+2)}\\ =\frac{{(k+1)}^2}{2(k+1)(k+2)}\\ =\frac{k+1}{2(k+2)}\end{array}$

Hence The formula is verified and The formula is $S_n=\frac{n}{2(n+1)}$