Mathematics - Sequence & Series Question with Solution | TestHub

$$\begin{array}{rl}& 1+x^4+x^5=\sum _{i=0}^5{a}_i(1+x{)}^i\\ & =a_0+a_1(1+x{)}^1+a_2(1+x{)}^2+a_3(1+x{)}^3\\ & +a_4(1+x{)}^4+a_5(1+x{)}^5\\ & \Rightarrow 1+x^4+x^5\end{array}$$ $$\begin{array}{rl}=a_0+a_1(1+x)& +a_2\left(1+2x+x^2\right)+a_3\left(1+3x+3x^2+x^3\right)\\ & +a_4\left(1+4x+6x^2+4x^3+x^4\right)\\ & +a_5\left(1+5x+10x^2+10x^3+5x^4+x^5\right)\\ \Rightarrow & 1+x^4+x^5\\ =a_0+a_1+a_{1}x+a_2+2a_{2}x+a_2{x}^2+a_3+3a_{3}x& \\ +3a_3{x}^2+a_3{x}^3+a_4+4a_{4}x+6a_4{x}^2+4a_4{x}^3+a_4{x}^4+a_5& \\ & +5a_{5}x+10a_5{x}^2+10a_5{x}^3+5a_5{x}^4+a_5{x}^5\end{array}$$ $$\begin{array}{rl}\Rightarrow & 1+x^4+x^5\\ =& \left(a_0+a_1+a_2+a_3+a_4+a_5\right)\\ & +x\left(a_1+2a_2+3a_3+4a_4+5a_5\right)\\ +x^2(a_2+& 3a_3+6a_4+10a_5)+x^3\left(a_3+4a_4+10a_5\right)\\ & +x^4\left(a_4+5a_5\right)+x^5\left(a_5\right)\end{array}$$ On comparing the like coefficients, we get $$\begin{array}{rl}& a_5=1\\ & a_4+5a_5=1\\ & a_3+4a_4+10a_5=0\end{array}$$ and $a_2+3a_3+6a_4+10a_5=0$ from (1) \& (2), we get $a_4=-4\dots (5)$ From (1), (3) \& (5), we get $a_3=+6$...(6) Now, from $(1),(5)$ and $(6)$, we get $a_2=-4$