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

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(D) 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.Sum $S = A \

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

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(D) 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.Sum $S = A \frac{1-r^n}{1-r} = (1+x)^{2016} \frac{1 - (\frac{x}{1+x})^{2017}}{1 - \frac{x}{1+x}} = (1+x)^{2016} \frac{\frac{(1+x)^{2017} - x^{2017}}{(1+x)^{2017}}}{\frac{1+x-x}{1+x}} = (1+x)^{2016} \frac{(1+x)^{2017} - x^{2017}}{(1+x)^{2017}} \cdot (1+x) = (1+x)^{2017} - x^{2017}$.We are given $\sum_{i=0}^{2016} a_i x^i = (1+x)^{2017} - x^{2017}$.Expanding $(1+x)^{2017}$ using the binomial theorem: $(1+x)^{2017} = \sum_{i=0}^{2017} {}^{2017}C_i x^i$.Thus, $\sum_{i=0}^{2016} a_i x^i = \left( \sum_{i=0}^{2017} {}^{2017}C_i x^i \right) - x^{2017} = \sum_{i=0}^{2016} {}^{2017}C_i x^i$.Comparing the coefficients of $x^{17}$, we get $a_{17} = {}^{2017}C_{17} = \frac{2017!}{17! (2017-17)!} = \frac{2017!}{17! 2000!}$.

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

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