Question

If ${A^2} = A$, then ${\left( {1 + A} \right)^4}$ is equal to
A. $I + A$
B. $I + 4A$
C. $I + 15A$
D. none of these

Answer 1

Hint: Whenever we have this type of problem, try to rewrite the problem which makes simplification easier. Now ${\left( {1 + A} \right)^4}$ can be written as ${\left( {1 + A} \right)^2}{\left( {1 + A} \right)^2}$. In this first find the value for ${\left( {1 + A} \right)^2}$ which is of the form ${(a + b)^2} = {a^2} + 2ab + {b^2}$. And then the result obtained will be multiplied twice to arrive at the correct answer.

Complete Step by Step Solution:
Here in this question we have given an expression which is ${\left( {1 + A} \right)^4}$ to simplify. First try to rewrite the given expression ${\left( {1 + A} \right)^4}$ in such a way that the simplification becomes easier.
Now ${\left( {1 + A} \right)^4}$ can be written as ${\left( {1 + A} \right)^2}{\left( {1 + A} \right)^2}$. If we find the value for ${\left( {1 + A} \right)^2}$ then we can easily get the answer for ${\left( {1 + A} \right)^4}$.
${\left( {1 + A} \right)^2}$ is of the form ${(a + b)^2}$ which can be solved using the formula given by: ${(a + b)^2} = {a^2} + 2ab + {b^2}$. Here $a = 1$ and $b = A$.
Now by making use of the ${(a + b)^2}$ formula we can find the value of ${\left( {1 + A} \right)^2}$ as below.
${\left( {1 + A} \right)^2} = \left( {1 + A} \right)\left( {1 + A} \right)$
On simplifying the above expression, we get
$ \Rightarrow {\left( {1 + A} \right)^2} = {1^2} + 2A + {A^2}$ or
$ \Rightarrow {\left( {1 + A} \right)^2} = 1 + 2A + {A^2}$
Now to find the value of ${\left( {1 + A} \right)^4}$ , we can write as
${\left( {1 + A} \right)^4} = {\left( {1 + A} \right)^2}{\left( {1 + A} \right)^2}$
$ \Rightarrow {\left( {1 + A} \right)^4} = \left( {1 + 2A + {A^2}} \right)\left( {1 + 2A + {A^2}} \right)$
Now simplify the above expression that is by multiplying the terms, we get
$ \Rightarrow {\left( {1 + A} \right)^4} = {1^2} + 2A + {A^2} + 2A + 4{A^2} + 2{A^3} + {A^2} + 2{A^3} + {A^4}$
Now, add the common terms, we get
$ \Rightarrow {\left( {1 + A} \right)^4} = 1 + 4A + 6{A^2} + 4{A^3} + {A^4}$
The above expression can be written as
$ \Rightarrow {\left( {1 + A} \right)^4} = 1 + 4A + 6{A^2} + 4{A^2}.A + {\left( {{A^2}} \right)^2}$
Now from the given question, we have ${A^2} = A$ , so wherever we have ${A^2}$ in the above expression replace it by $A$ . Therefore we get
$ \Rightarrow {\left( {1 + A} \right)^4} = 1 + 4A + 6A + 4A.A + {A^2}$
$ \Rightarrow {\left( {1 + A} \right)^4} = 1 + 4A + 6A + 4{A^2} + {A^2}$
Now, add the like terms, we get
$ \Rightarrow {\left( {1 + A} \right)^4} = 1 + 10A + 5{A^2}$
Again we have ${A^2}$ in the above expression replace it by $A$ , we get
$ \Rightarrow {\left( {1 + A} \right)^4} = 1 + 10A + 5A$
$ \Rightarrow {\left( {1 + A} \right)^4} = 1 + 15A$

Hence the option C is the correct answer.

Note:
Whenever we have this type of problems then try to reduce so that we can simplify easily otherwise if you know the ${\left( {a + b} \right)^4} = {a^4} + 4{a^3}b + 6{a^2}{b^2} + 4a{b^3} + {b^4}$ this formula then you can directly substitute and calculate the answer.