If 1+x^4+x^5=Σ_i=0^5 a_i (1+x^i ), for all x in R, then a_2 is:

$$\begin{aligned} &1+x^4+x^5=\sum_{i=0}^5 a_i(1+x)^i \\ &=a_0+a_1(1+x)^1+a_2(1+x)^2+a_3(1+x)^3 \\ &\quad+a_4(1+x)^4+a_5(1+x)^5 \\ &\Rightarrow 1+x^4+x^5 \end{aligned}$$​1+x4+x5=i=0∑5​ai​(1+x)i=a0​+a1​(1+x)1+a2​(1+x)2+a3​(1+x)3+a4​(1+x)4+a5​(1+x)5⇒1+x4+x5​ $$\begin{aligned} =a_0+a_1(1+x) &+a_2\left(1+2 x+x^2\right)+a_3\left(1+3 x+3 x^2+x^3\right) \\ &+a_4\left(1+4 x+6 x^2+4 x^3+x^4\right) \\ &+a_5\left(1+5 x+10 x^2+10 x^3+5 x^4+x^5\right) \\ \Rightarrow & 1+x^4+x^5 \\ =a_0+a_1+a_1 x+a_2+2 a_2 x+a_2 x^2+a_3+3 a_3 x \\ +3 a_3 x^2+a_3 x^3+a_4+4 a_4 x+6 a_4 x^2+4 a_4 x^3+a_4 x^4+a_5 \\ &+5 a_5 x+10 a_5 x^2+10 a_5 x^3+5 a_5 x^4+a_5 x^5 \end{aligned}$$=a0​+a1​(1+x)⇒=a0​+a1​+a1​x+a2​+2a2​x+a2​x2+a3​+3a3​x+3a3​x2+a3​x3+a4​+4a4​x+6a4​x2+4a4​x3+a4​x4+a5​​+a2​(1+2x+x2)+a3​(1+3x+3x2+x3)+a4​(1+4x+6x2+4x3+x4)+a5​(1+5x+10x2+10x3+5x4+x5)1+x4+x5+5a5​x+10a5​x2+10a5​x3+5a5​x4+a5​x5​ $$\begin{aligned} \Rightarrow & 1+x^4+x^5 \\ =&\left(a_0+a_1+a_2+a_3+a_4+a_5\right) \\ &+x\left(a_1+2 a_2+3 a_3+4 a_4+5 a_5\right) \\ +x^2\left(a_2+\right.&\left.3 a_3+6 a_4+10 a_5\right)+x^3\left(a_3+4 a_4+10 a_5\right) \\ &+x^4\left(a_4+5 a_5\right)+x^5\left(a_5\right) \end{aligned}$$⇒=+x2(a2​+​1+x4+x5(a0​+a1​+a2​+a3​+a4​+a5​)+x(a1​+2a2​+3a3​+4a4​+5a5​)3a3​+6a4​+10a5​)+x3(a3​+4a4​+10a5​)+x4(a4​+5a5​)+x5(a5​)​ On comparing the like coefficients, we get $$\begin{aligned} &a_5=1 \\ &a_4+5 a_5=1 \\ &a_3+4 a_4+10 a_5=0 \end{aligned}$$​a5​=1a4​+5a5​=1a3​+4a4​+10a5​=0​ and $a_2+3 a_3+6 a_4+10 a_5=0$a2​+3a3​+6a4​+10a5​=0 from (1) & (2), we get $a_4=-4 \ldots(5)$a4​=−4…(5) From (1), (3) & (5), we get $a_3=+6$a3​=+6...(6) Now, from $(1),(5)$(1),(5) and $(6)$(6), we get $a_2=-4$a2​=−4