Question

The coefficient of \[{{x}^{10}}\] in the expansion of \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}{{\left( 1+{{x}^{3}} \right)}^{4}}\] is equal to:
(a) 52
(b) 56
(c) 50
(d) 44

Answer 1

Hint: In order to find the solution of this question, we should have some knowledge of expansion like \[{{\left( a+b \right)}^{4}}={{a}^{4}}+4b{{a}^{3}}+6{{b}^{2}}{{a}^{2}}+4{{b}^{3}}a+{{b}^{4}}\], \[{{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\], \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]. By using these expansions we will be able to solve the question.

Complete Step-by-Step solution:
In this question, we have to find the coefficient of \[{{x}^{10}}\] in the expansion of \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}{{\left( 1+{{x}^{3}} \right)}^{4}}\]. To find the solution, we will start from expanding \[{{\left( 1+{{x}^{3}} \right)}^{4}}\] by using the formula of \[{{\left( a+b \right)}^{4}}\], that is \[{{\left( a+b \right)}^{4}}={{a}^{4}}+4b{{a}^{3}}+6{{b}^{2}}{{a}^{2}}+4{{b}^{3}}a+{{b}^{4}}\]. So, we can write \[{{\left( 1+{{x}^{3}} \right)}^{4}}\] as
\[{{\left( 1+{{x}^{3}} \right)}^{4}}={{\left( 1 \right)}^{4}}+4\left( {{x}^{3}} \right){{\left( 1 \right)}^{3}}+6{{\left( {{x}^{3}} \right)}^{2}}{{\left( 1 \right)}^{2}}+4{{\left( {{x}^{3}} \right)}^{3}}\left( 1 \right)+{{\left( {{x}^{3}} \right)}^{4}}\]
\[{{\left( 1+{{x}^{3}} \right)}^{4}}=1+4{{x}^{3}}+6{{x}^{6}}+4{{x}^{9}}+{{x}^{12}}\]
So, we can write \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}{{\left( 1+{{x}^{3}} \right)}^{4}}\] as
\[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\left( 1+4{{x}^{3}}+6{{x}^{6}}+4{{x}^{9}}+{{x}^{12}} \right)\]
Now, we are asked to find the coefficient of \[{{x}^{10}}\] in this expansion. So, with the terms of \[{{\left( 1+{{x}^{3}} \right)}^{4}}\] expansion, we will see the coefficients of what power of x is required to make the coefficients of x from \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\]. So, we can write,
Coefficient of \[{{x}^{10}}=\]
\[\begin{align}
  & 1\times \text{Coefficient of }{{x}^{10}}\text{ in }{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}+4\times \text{Coefficient of }{{x}^{7}}\text{ in }{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}+ \\
 & 6\times \text{Coefficient of }{{x}^{4}}\text{ in }{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}+4\times \text{Coefficient of }{{x}^{1}}\text{ in }{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}.....\left( i \right) \\
\end{align}\]
Here, we have ignored \[{{x}^{12}}\] because that is already larger than \[{{x}^{10}}\]. Now, we will find the coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\]. Here, we can see that the largest possible power of x is 8. So, the coefficient of \[{{x}^{10}}\] is not possible in \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\].
Now, we will find the coefficient of \[{{x}^{7}}\] in \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\] that is \[\left( 1+2x+{{x}^{2}} \right){{\left( 1+{{x}^{2}} \right)}^{3}}\] is equal to
\[2\times \text{Coefficent of }{{x}^{6}}\text{ in }{{\left( 1+{{x}^{2}} \right)}^{3}}+1\times \text{Coefficent of }{{x}^{5}}\text{ in }{{\left( 1+{{x}^{2}} \right)}^{3}}\]
And we know that, \[{{\left( 1+{{x}^{2}} \right)}^{3}}=1+3{{x}^{2}}+3{{x}^{4}}+{{x}^{6}}\]
So, we get a coefficient of \[{{x}^{6}}\] in \[{{\left( 1+{{x}^{2}} \right)}^{3}}\] as 1 and coefficient of \[{{x}^{5}}\] in \[{{\left( 1+{{x}^{2}} \right)}^{3}}\] as 0. Therefore, we get the coefficient of \[{{x}^{7}}\] in \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\] as
\[2\left( 1 \right)+0=2\]
That is, the coefficient of \[{{x}^{7}}\] in \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\] = 2.
Now, we will find the coefficient of \[{{x}^{4}}\] in \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\], that is \[\left( 1+2x+{{x}^{2}} \right)\left( 1+3{{x}^{2}}+3{{x}^{4}}+{{x}^{6}} \right)\] is equal to
\[1\times \text{Coefficent of }{{x}^{4}}\text{ in }{{\left( 1+{{x}^{2}} \right)}^{3}}+1\times \text{Coefficent of }{{x}^{2}}\text{ in }{{\left( 1+{{x}^{2}} \right)}^{3}}\]
\[1\times 3+2\times 3=3+3=6\]
That is, the coefficient of \[{{x}^{4}}\] in \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\] is 6.
Now, we will find the coefficient of x in \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\] that is \[\left( 1+2x+{{x}^{2}} \right)\left( 1+3{{x}^{2}}+3{{x}^{4}}+{{x}^{6}} \right)\] is equal to
\[2\times \text{constant of }{{\left( 1+{{x}^{2}} \right)}^{3}}\]
\[\Rightarrow 2\times 1=2\]
That is, the coefficient of x in \[{{\left( 1+x \right)}^{2}}{{\left( 1+{{x}^{2}} \right)}^{3}}\] is 2.
Now, we will put all these values in equation (i), so we will get,
Coefficient of \[{{x}^{10}}=1\times 0+4\times 2+6\times 6+4\times 2\]
Coefficient of \[{{x}^{10}}=8+36+8\]
Coefficient of \[{{x}^{10}}=52\]
Hence, option (a) is the right answer.

Note: While solving this question, one might think of expanding all the terms and then after multiplying and simplifying, find the coefficient of \[{{x}^{10}}\]. That is not the wrong way to do it, but it is a lengthier and complicated method to solve.