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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

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a

2016!16!

b

2017!2000!

c

2017!17!2000!

d

2016!17!1999!

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detailed solution

Correct option is C

we have , (1+x)2016+x(1+x)2015+x2(1+x)2014+…+x2016=∑i=02016 aixi ⇒(1+x)2016x1+x2017−1x1+x−1=∑i=12016 aixi⇒(1+x)2017−x2017=∑i=12016 aixi ⇒a17=Coefficient of x17 in (1+x)2017−x2017⇒a17=Coefficient of x17 in (1+x)2017⇒a17= 2017C17=2017!17!2000!

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