Question

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

Answer 1

Finding the coefficient:

Given equation, ${(1-x)}^{101}{(x^2+x+1)}^{100}$

$\begin{array}{rcl}& & {(1-x)}^{101}{(x^2+x+1)}^{100}\\ & =& (1-x){(1-x)}^{100}{(x^2+x+1)}^{100}\mathbf{\{}\mathbf{\because }{\mathit{a}}^{\mathbf{3}}\mathbf{-}{\mathit{b}}^{\mathbf{3}}\mathbf{=}\mathbf{(}\mathit{a}\mathbf{-}\mathit{b}\mathbf{\left)}\mathbf{\right(}{\mathit{a}}^{\mathbf{2}}\mathbf{+}\mathit{a}\mathit{b}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{)}\\ & =& (1-x){(1-x^3)}^{100}\end{array}$

The first expansion will not have a term of $x^{256}$. This is because $256$can't be divided by $3$.

The second expansion will have a term of $x^{256}$.

Therefore, the coefficient of $x^{256}$in ${(1-x)}^{101}{(x^2+x+1)}^{100}$will be in ${\left(\frac{255}{3}\right)}^{th}={85}^{th}$ term.

The coefficient is $=-(-C_{85}^{100})=C_{85}^{100}$

Therefore, the coefficient is $C_{85}^{100}$.