Chapter EP, Problem 11.7.5EP

To determine

State the proof of the given statement.

Explanation of Solution

Given:

The given statement is $\frac{5}{1\cdot 2}\cdot \frac{1}{3}+\frac{7}{2\cdot 3}\cdot \frac{1}{3^2}+\frac{9}{3\cdot 4}\cdot \frac{1}{3^3}+\cdot \cdot \cdot +\frac{2n+3}{n\left(n+1\right)}\cdot \frac{1}{3^n}=1-\frac{1}{3^n\left(n+1\right)}$ for all natural numbers.

Calculation:

Use mathematical induction.

  $S_1=1-\frac{1}{3^n\left(n+1\right)}=1-\frac{1}{3^1\left(1+1\right)}=1-\frac{1}{6}=\frac{5}{6}$

  $S_k=\frac{5}{1\cdot 2}\cdot \frac{1}{3}+\frac{7}{2\cdot 3}\cdot \frac{1}{3^2}+\frac{9}{3\cdot 4}\cdot \frac{1}{3^3}+\cdot \cdot \cdot +\frac{2k+3}{k\left(k+1\right)}\cdot \frac{1}{3^k}=1-\frac{1}{3^k\left(k+1\right)}$

  $S_{k+1}=\frac{5}{1\cdot 2}\cdot \frac{1}{3}+\frac{7}{2\cdot 3}\cdot \frac{1}{3^2}+\frac{9}{3\cdot 4}\cdot \frac{1}{3^3}+\cdot \cdot \cdot +\frac{2k+3}{k\left(k+1\right)}\cdot \frac{1}{3^k}+\frac{2\left(k+1\right)+3}{\left(k+1\right)\left(k+2\right)}\cdot \frac{1}{3^{k+1}}=1-\frac{1}{3^{k+1}\left(k+2\right)}$

Show that $S_k+\frac{2\left(k+1\right)+3}{\left(k+1\right)\left(k+2\right)}\cdot \frac{1}{3^{k+1}}=1-\frac{1}{3^{k+1}\left(k+2\right)}$

From L.H.S

  $S_k+\frac{2\left(k+1\right)+3}{\left(k+1\right)\left(k+2\right)}\cdot \frac{1}{3^{k+1}}$

  $\left[S_k=1-\frac{1}{3^k\left(k+1\right)}\right]$

  $\begin{array}{l}=1-\frac{1}{3^k\left(k+1\right)}+\frac{2\left(k+1\right)+3}{\left(k+1\right)\left(k+2\right)}\cdot \frac{1}{3^{k+1}}\\ =1-\frac{1}{3^{k+1}}\left[\frac{3}{\left(k+1\right)}-\frac{2\left(k+1\right)+3}{\left(k+1\right)\left(k+2\right)}\right]\\ =1-\frac{1}{3^{k+1}}\left[\frac{3\left(k+2\right)-\left(2k+5\right)}{\left(k+1\right)\left(k+2\right)}\right]\\ =1-\frac{1}{3^{k+1}}\left[\frac{3k+6-2k-5}{\left(k+1\right)\left(k+2\right)}\right]\\ =1-\frac{1}{3^{k+1}}\left[\frac{k+1}{\left(k+1\right)\left(k+2\right)}\right]\\ =1-\frac{1}{3^{k+1}}\left[\frac{1}{\left(k+2\right)}\right]\\ =1-\frac{1}{3^{k+1}\left(k+2\right)}=\text{R}\text{.H}\text{.S}\end{array}$

Hence the given statement is true for all natural numbers.