For x , x - 1 if ( 1 + x ) 2016 + x ( 1 + x ) 2015 + x 2 ( 1 + x ) 2014 + ? + x 2016 = i = 0 2016 a i x i then a 17 =

Question

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

Smart Achievers · Accepted Answer

(1)

$\sum_{i = 0}^{2016} a_{i} x^{i} = {(1 + x)}^{2017} - x^{2017}$

$a_{17} = coefficient of x^{17} in {(1 + x)}^{2017}$

Solution

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

1 $\frac{2017!}{17! 2000!}$

2 $\frac{2016!}{16!}$

3 $\frac{2016!}{17! 1999!}$

4 $\frac{2017!}{2000!}$

Ans.
(1)

$\sum_{i = 0}^{2016} a_{i} x^{i} = {(1 + x)}^{2017} - x^{2017}$

$a_{17} = coefficient of x^{17} in {(1 + x)}^{2017}$

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Solution

Q. For x∈ℝ, x≠-1 if (1+x)2016+x(1+x)2015+x2(1+x)2014+⋯+x2016=∑i=02016aixi then a17 =

1 2017!17!2000!

2 2016!16!

3 2016!17!1999!

4 2017!2000!

Ans. (1) ∑i=02016aixi=(1+x)2017-x2017 a17=coefficient of x17 in (1+x)2017