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If $$1 + x^4 + x^5 = \sum_{i=0}^{5} a_i(1+x)^i$$, for all $$x$$ in R, then $$a_2$$ is:
Let $$y = 1+x$$, which implies $$x = y-1$$.
The equation becomes $$1 + (y-1)^4 + (y-1)^5 = \sum_{i=0}^{5} a_i y^i$$
Since the RHS is a polynomial in $$y$$, $$a_2$$ is specifically the coefficient of the $$y^2$$ term on the LHS.
The $$y^2$$ term in $$(y-1)^4$$ is $$\binom{4}{2}y^2(-1)^2 = 6y^2$$. The $$y^2$$ term in $$(y-1)^5$$ is $$\binom{5}{2}y^2(-1)^3 = -10y^2$$.
$$a_2 = 6 - 10 = \mathbf{-4}$$
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