A) \[\frac{2016!}{16!}\]
B) \[\frac{2017!}{2000!}\]
C) \[\frac{2017!}{17!2000!}\]
D) \[\frac{2016!}{17!1999!}\]
Correct Answer: C
Solution :
\[\sum\limits_{i=0}^{2016}{{{c}_{i}}.{{x}^{i}}={{(1+x)}^{2016}}+x{{(1+x)}^{2016}}+{{x}^{2}}{{(1+x)}^{2014}}}\]\[+..........+{{x}^{2016}}\] \[=\frac{{{(1+x)}^{2016}}\left( 1-{{\left( \frac{x}{1+x} \right)}^{2017}} \right)}{1-\frac{x}{1+x}}\] \[\frac{=\frac{{{(1+x)}^{2016}}}{1}-\frac{{{x}^{2017}}}{(1+x)}}{\frac{x+1-x}{1+x}}=\frac{{{(1+x)}^{2017}}-{{x}^{2017}}}{1}\] \[\therefore \]\[{{a}_{17}}{{=}^{2017}}{{C}_{17}}=\frac{2007!}{17!2000!}\]
You need to login to perform this action.
You will be redirected in
3 sec
