Let (x + 10)^{50} + (x - 10)^{50} = a_0 + a_1 | vedclass

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(C) Given the expansion: $(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1x + a_2x^2 + .... + a_{50}x^{50}$.To find $a_0$,set $x = 0$:$a_0 = (0 + 10)^{50} +

Vedclass · Accepted Answer

$12.25$

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(C) Given the expansion: $(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1x + a_2x^2 + .... + a_{50}x^{50}$.To find $a_0$,set $x = 0$:$a_0 = (0 + 10)^{50} + (0 - 10)^{50} = 10^{50} + 10^{50} = 2 	imes 10^{50}$.To find $a_2$,we look for the coefficient of $x^2$ in the expansion of $(x + 10)^{50} + (x - 10)^{50}$.Using the binomial theorem,$(x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^k a^{n-k}$.For $(x + 10)^{50}$,the term with $x^2$ is $\binom{50}{2} x^2 (10)^{48}$.For $(x - 10)^{50}$,the term with $x^2$ is $\binom{50}{2} x^2 (-10)^{48} = \binom{50}{2} x^2 (10)^{48}$.Thus,$a_2 = \binom{50}{2} (10)^{48} + \binom{50}{2} (10)^{48} = 2 	imes \binom{50}{2} 	imes 10^{48}$.Now,calculate the ratio $\frac{a_2}{a_0}$:$\frac{a_2}{a_0} = \frac{2 	imes \binom{50}{2} 	imes 10^{48}}{2 	imes 10^{50}} = \frac{\binom{50}{2}}{10^2} = \frac{\frac{50 	imes 49}{2}}{100} = \frac{1225}{100} = 12.25$.

Let $(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1x + a_2x^2 + .... + a_{50}x^{50}$,for $x \in R$; then $\frac{a_2}{a_0}$ is equal to

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