(x1+x)2017š•’µ

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Question:

For x āˆˆ R, x ā‰  -1, if (1+x)^2016 + x(1+x)^2015 + x^2(1+x)^2014 + ... + x^2016 = 2016āˆ‘_(i=0)^(2016) a_i x^i, then a_17 is equal to:

2017! / (17! 2000!)

2016! / (17! 1999!)

2016! / 2000!

2016! / 16!

Solution:

It is a GP with(1+x)2016as 1st term andx1+xas common ratio.Total number of terms are2017Sum=a(rnš•’µ)rš•’µ=(1+x)2016(x1+x)2017š•’µ