Question

a) Find a formula for

$\frac{\mathbf{1}}{\mathbf{1}\mathbf{\cdot }\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\cdot }\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}$

by examining the values of this expression for small

values of n.

b) Prove the formula you conjectured in part (a).

Short answer

(a) The formula is,$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{n(n+1)}=\frac{n}{n+1}$

(b) The formula $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{n(n+1)}=\frac{n}{n+1}$is proved.

Step-by-step solution

Principle of Mathematical Induction

To prove that P(n)is true for all positive integers n, where p(n)is a propositional function, we complete two steps:

Basis Step:

We verify that p(1)is true.

Inductive Step:

We show that the conditional statement $\mathit{P}\left(k\right)\mathbf{\to }\mathit{P}\left(k+1\right)$is true for all positive integers k.

(a)Step 2:  Finding the formula

Given expression is

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{n(n+1)}$

For n = 1

$\frac{1}{1.2}=\frac{1}{2}$

For n = 2

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}=\frac{2}{(2+1)}$

For n = 3

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}=\frac{3}{(3+1)}$

From the above we can conclude that

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{n(n+1)}=\frac{n}{n+1}$

(b) Step 3: Proving the formula

We will prove the formula using the principle of mathematics induction

Proving Basis step

The statement which we have to prove is

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{n(n+1)}=\frac{n}{n+1}$

In the basis step, we need to prove that P(1) is true

For finding statement P(1) substituting 1 for n in the statement

Therefore, the statement P(1) is

$\begin{array}{r}\frac{1}{1\cdot 2}=\frac{1}{1+1}\\ \frac{1}{2}=\frac{1}{2}\end{array}$

Therefore, the statement P(1) is true this is also known as the basis step of the proof.

Proving the Inductive step

In the inductive step, we need to prove that, if P(k) is true, then P(K+1) is also true.

That is,

$P\left(k\right)\to P\left(k+1\right)$is true for all positive integers k.

In the inductive hypothesis, we assume that P(k) is true for any arbitrary positive integer

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{k(k+1)}=\frac{k}{k+1}$...(i)

Now we must have to show that is also true

Therefore replacing with in the statement

$\begin{array}{r}\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{(k+1)\cdot \left(\right(k+1)+1)}=\frac{(k+1)}{(k+1)+1}\\ =\frac{k+1}{k+2}\end{array}$

Now, Adding $\frac{1}{(k+1)\cdot \left(\right(k+1)+1)}$ in both sides of the equation (i) or inductive hypothesis.

$\begin{array}{r}\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{k(k+1)}+\frac{1}{(k+1)\cdot \left(\right(k+1)+1)}=\frac{k}{k+1}+\frac{1}{(k+1)\cdot \left(\right(k+1)+1)}\\ =\frac{1}{(k+1)}\left[k+\frac{1}{(k+2)}\right]\\ =\frac{1}{(k+1)}\left[\frac{k^2+2k+1}{(k+2)}\right]\\ =\frac{1}{(k+1)}\left[\frac{(k+1{)}^2}{(k+2)}\right]\\ =\frac{k+1}{k+2}\end{array}$

From the above, we can see that P(k+1) is also true

Hence, P(k+1)is true under the assumption that P(k)is true. This

completes the inductive step.

Hence the formula$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{n(n+1)}=\frac{n}{n+1}$is proved.