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

Question

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

3 Answers

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

Your answer

3 Answers

Your comment on this answer:

Your comment on this answer:

Your comment on this answer:

Related questions

Categories