Prove by mathematical induction $\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{n\left( {n + 1} \right)}} = \dfrac{n}{{n + 1}}$

Hint: The given equation is $\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{n\left( {n + 1} \right)}} = \dfrac{n}{{n + 1}}$
Here, we need to prove the given equation by mathematical induction. We use mathematical induction to prove a statement is true for all natural numbers. We need to deal with this method using two steps.
Formula to be used:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$

Complete answer:
Mathematical induction is a method that is used to prove a statement, or a theorem is true for all the natural numbers. This method involves two steps which are given below.
Step \[1\]-First we shall consider an initial value. We need to prove that the given statement is true for that initial value. This is the base step of the induction method.
Step $2$- Now, we need to assume that the statement is true for any value $n = k$. Then we need to prove the statement is true for $n = k + 1$. This step is usually known as the inductive step.
The given equation is $\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{n\left( {n + 1} \right)}} = \dfrac{n}{{n + 1}}$
Step \[1\]- Let us consider an initial value $n = 1$.
We shall substitute $n = 1$in the given statement.
$\dfrac{1}{{1.2}} = \dfrac{1}{{1 + 1}}$
$ \Rightarrow \dfrac{1}{2} = \dfrac{1}{2}$
Hence, the given statement is true for the initial value $n = 1$
Step $2$- Now, we need to assume that the statement is true for any value $n = k$.
That means we have $\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{k\left( {k + 1} \right)}} = \dfrac{k}{{k + 1}}$ ………. $\left( 1 \right)$
Then we need to prove the statement is true for $n = k + 1$.
Let us consider the left-hand side of the statement.
$\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{k + 1\left( {k + 1 + 1} \right)}}$ $ = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{k\left( {k + 1} \right)}} + \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$
(Now, we need to substitute the equation $\left( 1 \right)$)
$ = \dfrac{k}{{k + 1}} + \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$
$ = \dfrac{{k\left( {k + 2} \right) + 1}}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$
$ = \dfrac{{{k^2} + 2k + 1}}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$ (Here we have applied the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$)
$ = \dfrac{{{{\left( {k + 1} \right)}^2}}}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$
$ = \dfrac{{\left( {k + 1} \right) \times \left( {k + 1} \right)}}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$
$ = \dfrac{{k + 1}}{{k + 2}}$
Now, let us consider the right-hand side of the statement.
$\dfrac{{k + 1}}{{k + 1 + 1}}$ $ = \dfrac{{k + 1}}{{k + 2}}$
Hence, the left-hand side and the right-hand side of the statement are the same for $n = k + 1$.
Thus, the given statement is true for $n = k + 1$ assuming it is true for $n = k$.
Thus, $\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{n\left( {n + 1} \right)}} = \dfrac{n}{{n + 1}}$

Note:
Generally, the mathematical induction method is applied only for natural numbers. Here, the base step is to prove the statement is true for the initial value and the inductive step is to prove the statement is true for $n = k + 1$assuming it is true for $n = k$. Then, we can able to say that the statement is true for all natural numbers.