Expand ${\left( {99} \right)^2}$ using suitable identities.

Hint: In this question use the identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ and write 99 as (100 – 1) so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given expression
${\left( {99} \right)^2}$
Now we have to expand this using suitable identities.
$ \Rightarrow {\left( {99} \right)^2} = {\left( {100 - 1} \right)^2}$
Now we all know the identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ so, use this identity we have,
a=100 ,b=1
$ \Rightarrow {\left( {99} \right)^2}=100^2 + 1^2-2(100)(1)$
$ \Rightarrow {\left( {99} \right)^2} = 10000 + 1 - 200$
Now it is a simple addition so we have,
$ \Rightarrow {\left( {99} \right)^2} = 9801$
So this is the required answer.

Note – In such types of questions the key concept we have to remember is the basic general identity of square (i.e. ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$) so apply this property as above and simplify, we can use another method to expand $(99)^2$ which is by using Binomial theorem of expansion of ${(1 + x)}^{n}$ which is given as ${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ..............$