For x in R, x neq -1, if (1 + x)^{2016} + x(1 | vedclass

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(A) The given expression is a geometric series with first term $A = (1 + x)^{2016},$ common ratio $r = \frac{x}{1 + x},$ and $n = 2017$ terms.Using th

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$\frac{2017!}{17! 2000!}$

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(A) The given expression is a geometric series with first term $A = (1 + x)^{2016},$ common ratio $r = \frac{x}{1 + x},$ and $n = 2017$ terms.Using the sum formula for a geometric series $S = A \frac{1 - r^n}{1 - r}:$$S = (1 + x)^{2016} \frac{1 - (\frac{x}{1 + x})^{2017}}{1 - \frac{x}{1 + x}}$$S = (1 + x)^{2016} \frac{1 - \frac{x^{2017}}{(1 + x)^{2017}}}{\frac{1 + x - x}{1 + x}}$$S = (1 + x)^{2016} \frac{\frac{(1 + x)^{2017} - x^{2017}}{(1 + x)^{2017}}}{\frac{1}{1 + x}}$$S = (1 + x)^{2016} \frac{(1 + x)^{2017} - x^{2017}}{(1 + x)^{2017}} \cdot (1 + x)$$S = (1 + x)^{2017} - x^{2017}$We need the coefficient of $x^{17}$ in the expansion of $(1 + x)^{2017} - x^{2017}.$Using the Binomial Theorem,$(1 + x)^{2017} = \sum_{k=0}^{2017} \binom{2017}{k} x^k.$The coefficient of $x^{17}$ is $\binom{2017}{17} = \frac{2017!}{17! 2000!}.$

For $x \in R, x \neq -1,$ if $(1 + x)^{2016} + x(1 + x)^{2015} + x^2(1 + x)^{2014} + \dots + x^{2016} = \sum_{i = 0}^{2016} a_i x^i,$ then $a_{17}$ is equal to

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