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

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

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kritika · Answer 1 · 2021-12-26T04:52:46+0000

Ans: (D)
Hint:

$a_{17}=$ coeff

$^{n}$ of

$x^{17}$

$\begin{aligned} &={ }^{2016} \mathrm{C}_{17}+{ }^{2015} \mathrm{C}_{16}+{ }^{2014} \mathrm{C}_{15}+\cdots+{ }^{1999} \mathrm{C}_{0} \\ &={ }^{2016} \mathrm{C}_{1999}+{ }^{2015} \mathrm{C}_{1999}+{ }^{2014} \mathrm{C}_{1999}+\cdots+{ }^{1999} \mathrm{C}_{1999} \\ &={ }^{2017} \mathrm{C}_{2000}=\frac{\mid 2017}{\mid 17 {2000!}{}} \end{aligned}$

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

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