First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

$\frac{2016!}{16!}$
B

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

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

$\frac{2016!}{17! 1999!}$

Solution

we have ,
$(1 + x)^{2016} + x (1 + x)^{2015} + x^{2} (1 + x)^{2014} + \dots + x^{2016} = \sum_{i = 0}^{2016} a_{i} x^{i}$
$\Rightarrow (1 + x)^{2016} (\frac{{(\frac{x}{1 + x})}^{2017} - 1}{\frac{x}{1 + x} - 1}) = \sum_{i = 1}^{2016} a_{i} x^{i}$
$\Rightarrow \{(1 + x)^{2017} - x^{2017}\} = \sum_{i = 1}^{2016} a_{i} x^{i}$
$\Rightarrow a_{17} =$ Coefficient of $x^{17}$ in $\{(1 + x)^{2017} - x^{2017}\}$
$\Rightarrow a_{17} =$ Coefficient of $x^{17}$ in $(1 + x)^{2017}$
$\Rightarrow a_{17} =$ $^{2017} C_{17} = \frac{2017!}{17! 2000!}$

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