If $1+x^4+x^5=\sum_{i=0}^5 a_i\left(1+x^i\right)$, for all $x$ in $R$, then $a_2$ is:

Question

Aicharya · Accepted Answer

$ \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} $ $ \begin{aligned} =a_0+a_1(1+x) &+a_2\left(1+2 x+x^2ight)+a_3\left(1+3 x+3 x^2+x^3ight) \ &+a_4\left(1+4 x+6 x^2+4 x^3+x^4ight) \ &+a_5\left(1+5 x+10 x^2+10 x^3+5 x^4+x^5ight) \ \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} $ $ \begin{aligned} \Rightarrow & 1+x^4+x^5 \ =&\left(a_0+a_1+a_2+a_3+a_4+a_5ight) \ &+x\left(a_1+2 a_2+3 a_3+4 a_4+5 a_5ight) \ +x^2\left(a_2+ight.&\left.3 a_3+6 a_4+10 a_5ight)+x^3\left(a_3+4 a_4+10 a_5ight) \ &+x^4\left(a_4+5 a_5ight)+x^5\left(a_5ight) \end{aligned} $ 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} $ and $a_2+3 a_3+6 a_4+10 a_5=0$ from (1) \& (2), we get $a_4=-4 \ldots(5)$ From (1), (3) \& (5), we get $a_3=+6$...(6) Now, from $(1),(5)$ and $(6)$, we get $a_2=-4$

If $1+x^4+x^5=\sum_{i=0}^5 a_i\left(1+x^i\right)$, for all $x$ in $R$, then $a_2$ is:

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