Question

Find the coefficient of ${{x}^{256}}$in ${{(1-x)}^{101}}{{({{x}^{2}}+x+1)}^{100}}$

Answer 1

Hint: This question is based on binomial expansion. First we write ${{(1-x)}^{101}}$ as the product of ( 1-x) and ${{(1-x)}^{100}}$. Our expression will become ( 1 – x) ${{(1-x)}^{100}}$${{({{x}^{2}}+x+1)}^{100}}$. Now expression becomes ${{(1-{{x}^{3}})}^{100}}-x{{(1-{{x}^{3}})}^{100}}$. The coefficient of ${{x}^{100}}$is equal to the summation of the coefficient of ${{x}^{100}}$from ${{(1-{{x}^{3}})}^{100}}$and $x{{(1-{{x}^{3}})}^{100}}$. Expand the expression ${{(1-{{x}^{3}})}^{100}}$by using binomial expression formula ${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{n}}$. As we see that the exponent of x is a multiple of 3. For the coefficient of ${{x}^{256}}$ in ${{(1-{{x}^{3}})}^{100}}$ is -1. By solving this, we get our desirable answer.

Complete step by step solution: 
We are given an expression and we have to find the coefficient of ${{x}^{256}}$
Expression is ${{(1-x)}^{101}}{{({{x}^{2}}+x+1)}^{100}}$----------- (1)
We can write ${{(1-x)}^{101}}$in the form of ( 1 – x ) ${{(1-x)}^{100}}$
That is ${{(1-x)}^{101}}$= ( 1 – x ) ${{(1-x)}^{100}}$----------- (2)
Now by putting equation (2) in equation (1), we get
${{(1-x)}^{101}}{{({{x}^{2}}+x+1)}^{100}}$= (1-x) ${{(1-x)}^{100}}$${{({{x}^{2}}+x+1)}^{100}}$
On simplifying the above equation, we get
$\Rightarrow (1-x){{\left\{ \left( 1-x \right)\left( {{x}^{2}}+x+1 \right) \right\}}^{100}}$
$\Rightarrow (1-x){{\left\{ \left( 1+x+{{x}^{2}} \right)-x\left( {{x}^{2}}+x+1 \right) \right\}}^{100}}$
${}^{100}{{C}_{1}}({{x}^{3}})+{}^{100}{{C}_{2}}{{({{x}^{3}})}^{2}}+{}^{100}{{C}_{3}}{{({{x}^{3}})}^{3}}+.......{}^{100}{{C}_{n}}{{({{x}^{3}})}^{n}}$$\Rightarrow (1-x){{\{(1+x+{{x}^{2}}-{{x}^{3}}-{{x}^{2}}-x)\}}^{100}}$
$\Rightarrow (1-x){{\{(1-{{x}^{3}})\}}^{100}}$
$\Rightarrow {{(1-{{x}^{3}})}^{100}}-x{{(1-{{x}^{3}})}^{100}}$
We have to find the coefficient of ${{x}^{256}}$in the above expression.
For the coefficient of ${{x}^{256}}$, we have to find the summation of the coefficients of ${{x}^{256}}$from ${{(1-{{x}^{3}})}^{100}}$and $x{{(1-{{x}^{3}})}^{100}}$.
The expansion of ${{(1-{{x}^{3}})}^{100}}$is given by
${{(1-{{x}^{3}})}^{100}}$ = 1 + ${}^{100}{{C}_{1}}({{x}^{3}})+{}^{100}{{C}_{2}}{{({{x}^{3}})}^{2}}+{}^{100}{{C}_{3}}{{({{x}^{3}})}^{3}}+.......{}^{100}{{C}_{n}}{{({{x}^{3}})}^{n}}$
Note that, here we have only those powers of x that are multiples of 3.
Since 85 is a multiple of 3, therefore coefficient of${{x}^{256}}$ in the expansion ${{(1-{{x}^{3}})}^{100}}$is -1.
First term will not have ${{x}^{256}}$
Coefficient of ${{x}^{256}}$is -1 $\times $coefficient of ${{x}^{256}}$in ${{(1-{{x}^{3}})}^{100}}$
-1 (- ${}^{100}{{C}_{85}}$)
= ${}^{100}{{C}_{85}}$
Hence the coefficient of ${{x}^{256}}$in ${{(1-x)}^{101}}{{({{x}^{2}}+x+1)}^{100}}$is ${}^{100}{{C}_{85}}$

Note: In this question, we should try to simply expand the expression by using the binomial expansion formula. Most of the students made mistakes in expanding the formula. We should be careful while expanding it.