For x ∈R, x≠-1, if 1+x^2016+x1+x^2015+x^21+x^2014+…+x^2016= ∑ i=0 2016 a_i x_i , then a_17 is equal to

Question

MARKS Content Team · Accepted Answer

Correct answer: (A) (2017!)/(17!2000!). S=1+x^2016+x1+x^2015+x^21+x^2014+……+x^20151+x+x^2016 ...i x/(1+x) S=x 1+x^2015+x^21+x^2014+…+x^2016+x^20171+x ...ii ii-i S/(1+x)=1+x^2016-x^20171+x ∴ S=1+x^2017-x^2017 Coefficient a_17=^2017C_17 =(2017!)/(17!2000!)

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 $a_{17}$ is equal to

Solution

Relevant Questions

For x ∈R, x≠-1, if 1+x2016+x1+x2015+x21+x2014+…+x2016= ∑ i=0 2016 a i x i , then a17 is equal to

Solution

Relevant Questions

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 $a_{17}$ is equal to