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7.Binomial Theorem
hard
For $x\, \in \,R\,,\,x\, \ne \, - 1,$ if ${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ....{x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i\,}{\,x^i}} ,$ then $a_{17}$ is equal to
A
$\frac{{2017\,!\,}}{{17\,!\,2000\,!}}$
B
$\frac{{2016\,!\,}}{{17\,!\,1999\,!}}$
C
$\frac{{2016\,!\,}}{{16\,!}}$
D
$\frac{{2017\,!\,}}{{2000\,!}}$
(JEE MAIN-2016)
Solution
$S=(1+x)^{2016}+x(1+x)^{2015}+x^{2}(1+x)^{2014}$
$+\ldots+x^{2015}(1+x)+x^{2016}……..(i)$
$\left(\frac{x}{1+x}\right) S=x(1+x)^{2015}+x^{2}(1+x)^{2014}$
$+\ldots +x^{2016}+\frac{x^{2017}}{1+x}……..(ii)$
Subtracting $(i)$ from $(ii)$
$\frac{S}{1+x}=(1+x)^{2016}-\frac{x^{2017}}{1+x}$
$\therefore \quad S=(1+x)^{2017}-x^{2017}$
$a_{17}=$ coefficient of $x^{17}=^{2017} C_{17}$
$=\frac{2017 !}{17 ! 2000 !}$
Standard 11
Mathematics
