If 1 + x^4 + x^5 = sumlimits_{i = 0}^5 a_i (1 | vedclass

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(A) Given $1 + x^4 + x^5 = \sum\limits_{i = 0}^5 a_i (1 + x)^i.$Let $y = 1 + x$,so $x = y - 1$.Substituting this into the expression:$1 + (y - 1)^4 +

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$-4$

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(A) Given $1 + x^4 + x^5 = \sum\limits_{i = 0}^5 a_i (1 + x)^i.$Let $y = 1 + x$,so $x = y - 1$.Substituting this into the expression:$1 + (y - 1)^4 + (y - 1)^5 = \sum\limits_{i = 0}^5 a_i y^i.$Expanding the terms using the Binomial Theorem:$(y - 1)^4 = y^4 - 4y^3 + 6y^2 - 4y + 1$$(y - 1)^5 = y^5 - 5y^4 + 10y^3 - 10y^2 + 5y - 1$Adding these to $1$:$1 + (y^4 - 4y^3 + 6y^2 - 4y + 1) + (y^5 - 5y^4 + 10y^3 - 10y^2 + 5y - 1) = y^5 - 4y^4 + 6y^3 - 4y^2 + y + 1.$Comparing this with $\sum\limits_{i = 0}^5 a_i y^i = a_5 y^5 + a_4 y^4 + a_3 y^3 + a_2 y^2 + a_1 y + a_0$,we get:$a_5 = 1, a_4 = -4, a_3 = 6, a_2 = -4, a_1 = 1, a_0 = 1.$Thus,$a_2 = -4$.

If $1 + x^4 + x^5 = \sum\limits_{i = 0}^5 a_i (1 + x)^i$ for all $x$ in $\mathbb{R},$ then $a_2$ is

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