Chapter 9.4, Problem 43E

To determine

To solve:Find a formula for the sum of the first n terms of the sequence $\frac{1}{4},\frac{1}{12},\frac{1}{24},\frac{1}{40},\mathrm{.....},\frac{1}{2n(n+1)}$ and verify it.

Answer to Problem 43E

  $S=\frac{n}{2(n+1)}$

Explanation of Solution

Given:

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

Calculation:

The given sequence $\frac{1}{4},\frac{1}{12},\frac{1}{24},\frac{1}{40},\mathrm{.....},\frac{1}{2n(n+1)}$ can be expressed as:

  $\begin{array}{l}\frac{1}{2(1+1)},\frac{1}{4(2+1)},\frac{1}{6(3+1)},\frac{1}{8(4+1)},\mathrm{.....},\frac{1}{2n(n+1)}\\ =\frac{1}{2\times 2},\frac{1}{4\times 3},\frac{1}{6\times 4},\frac{1}{8\times 5},\mathrm{.....},\frac{1}{2n(n+1)}\end{array}$

The sum of the series is given by

  $\begin{array}{l}S=\frac{1}{2\times 2}+\frac{1}{4\times 3}+\frac{1}{6\times 4}+\frac{1}{8\times 5}+\mathrm{.....}+\frac{1}{2n(n+1)}\\ =>S=\frac{1}{2}\left(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+\mathrm{.....}+\frac{1}{n(n+1)}\right)\\ =>S=\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\mathrm{.......}+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right]\\ =>S=\frac{1}{2}\left[1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{5}-\frac{1}{5}\right)-\mathrm{.......}-\left(\frac{1}{n}-\frac{1}{n}\right)-\frac{1}{n+1}\right]\\ =>S=\frac{1}{2}\left[1-\frac{1}{n+1}\right]\\ =>S=\frac{n}{2(n+1)}------------(1)\end{array}$

To verify our result, let us find the sum of the first two terms of the sequence manually and also by using the formula and examine if both are same or not.

Sum of first two terms $=\left(\frac{1}{4}+\frac{1}{12}\right)=\frac{3+1}{12}=\frac{1}{3}$

Now, in (1) putting $n=2$ we get:

  $=\frac{2}{2(2+1)}=\frac{1}{3}$

Hence, our formula is verified.