A) \[{{2}^{16}}\]
B) \[{{2}^{15}}\]
C) \[{{2}^{14}}\]
D) None of these
Correct Answer: C
Solution :
We have \[^{15}{{C}_{0}}+{{\,}^{15}}{{C}_{1}}+...+{{\,}^{15}}{{C}_{15}}={{2}^{15}}\] \[\Rightarrow \,\,\,2{{(}^{15}}{{C}_{8}}+{{\,}^{15}}{{C}_{9}}+...+{{\,}^{15}}{{C}_{15}})={{2}^{15}}\] \[(\because \,{{\,}^{n}}{{C}_{r}}={{\,}^{n}}{{C}_{n-r}})\] Þ \[^{15}{{C}_{8}}+{{\,}^{15}}{{C}_{9}}+...+{{\,}^{15}}{{C}_{15}}={{2}^{14}}\]You need to login to perform this action.
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