Question

The coefficient of ${{x}^{4}}$ in the expansion of ${{\left( 1+x+{{x}^{2}}+{{x}^{3}} \right)}^{11}}$is:
A.900
B.909
C.990
D.None of these

Answer 1

Hint: Here we have to find the coefficient of ${{x}^{4}}$in the expansion of${{\left( 1+x+{{x}^{2}}+{{x}^{3}} \right)}^{11}}$. For that, we will first factorize the equation inside the bracket. It will get split into two terms and then we will expand the two terms separately and then we will multiply the coefficients of those powers which can give the term${{x}^{4}}$and then we will add them. The required answer will be the coefficient of ${{x}^{4}}$.

Complete step by step solution:
The given expression is ${{\left( 1+x+{{x}^{2}}+{{x}^{3}} \right)}^{11}}$. We will first factorize the function given in the bracket i.e. we will factorize $1+x+{{x}^{2}}+{{x}^{3}}$
$\Rightarrow 1+x+{{x}^{2}}+{{x}^{3}}=\left( 1+x \right)+{{x}^{2}}\left( 1+x \right)=\left( 1+{{x}^{2}} \right)\left( 1+x \right)$
We will put this in the given expression ${{\left( 1+x+{{x}^{2}}+{{x}^{3}} \right)}^{11}}$
$\Rightarrow$ ${{\left( 1+x+{{x}^{2}}+{{x}^{3}} \right)}^{11}}={{\left( 1+x \right)}^{11}}{{\left( 1+{{x}^{2}} \right)}^{11}}$
Now, we will first expand the term ${{\left( 1+x \right)}^{11}}$.
${{\left( 1+x \right)}^{11}}={}^{11}{{C}_{0}}{{\left( 1 \right)}^{11}}{{\left( x \right)}^{0}}+{}^{11}{{C}_{1}}{{\left( 1 \right)}^{10}}{{\left( x \right)}^{1}}+{}^{11}{{C}_{2}}{{\left( 1 \right)}^{9}}{{\left( x \right)}^{2}}+{}^{11}{{C}_{3}}{{\left( 1 \right)}^{8}}{{\left( x \right)}^{3}}+{}^{11}{{C}_{4}}{{\left( 1 \right)}^{7}}{{\left( x \right)}^{4}}+..........+{}^{11}{{C}_{11}}{{\left( 1 \right)}^{0}}{{\left( x \right)}^{11}}$ On further simplifying the terms, we get
$\Rightarrow$ ${{\left( 1+x \right)}^{11}}=1+11x+55{{x}^{2}}+165{{x}^{3}}+330{{x}^{4}}+..........+{{x}^{11}}$…………….$\left( 1 \right)$
Now, we will first expand the term${{\left( 1+{{x}^{2}} \right)}^{11}}$.
${{\left( 1+{{x}^{2}} \right)}^{11}}={}^{11}{{C}_{0}}{{\left( 1 \right)}^{11}}{{\left( {{x}^{2}} \right)}^{0}}+{}^{11}{{C}_{1}}{{\left( 1 \right)}^{10}}{{\left( {{x}^{2}} \right)}^{1}}+{}^{11}{{C}_{2}}{{\left( 1 \right)}^{9}}{{\left( {{x}^{2}} \right)}^{2}}+{}^{11}{{C}_{3}}{{\left( 1 \right)}^{8}}{{\left( {{x}^{2}} \right)}^{3}}+..........+{}^{11}{{C}_{11}}{{\left( 1 \right)}^{0}}{{\left( {{x}^{2}} \right)}^{11}}$ On further simplifying the terms, we get
$\Rightarrow$ ${{\left( 1+{{x}^{2}} \right)}^{11}}=1+11{{x}^{2}}+55{{x}^{4}}+165{{x}^{6}}+..........+{{x}^{22}}$ ………………….$\left( 2 \right)$
Now, we will multiply the powers of those powers which will give the term.
$\Rightarrow$ ${{\left( 1+{{x}^{2}} \right)}^{11}}=55{{x}^{2}}.11{{x}^{2}}+1.330{{x}^{4}}+55{{x}^{4}}.1$
On simplifying the terms further, we get
$\Rightarrow$ ${{\left( 1+{{x}^{2}} \right)}^{11}} =605{{x}^{4}}+330{{x}^{4}}+55{{x}^{4}}$
 On addition, we get
$\Rightarrow$ ${{\left( 1+{{x}^{2}} \right)}^{11}} =990{{x}^{4}}$
Thus, the coefficient of ${{x}^{4}}$ is 990.
Hence, the correct answer is option C.

Note: Here we have calculated the value of terms like${}^{n}{{C}_{m}}=\dfrac{n!}{m!\times \left( n-m \right)!}$, this is the ratio of factorial of the terms.
We need to know the meaning of the factorial for solving problems like that.
1.Factorial of any positive integer is defined as the multiplication of all the positive integers less than or equal to the given positive integers.
2.Factorial of zero is one.
3.Factorials are commonly used in permutations and combinations problems.
4.Factorials of negative integers are not defined