A) \[{{2}^{n-1}}\]
B) \[{{2}^{n-1}}\]
C) \[{{2}^{n}}\]
D) \[{{2}^{n-1}}-1\]
Correct Answer: A
Solution :
\[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+....+{{C}_{n}}{{x}^{n}}\] Putting x = 1, we get \[\Rightarrow {{2}^{n}}={{C}_{0}}+{{C}_{1}}+{{C}_{2}}+.....+{{C}_{n}}\] .....(i) or \[{{C}_{1}}+{{C}_{2}}+{{C}_{3}}+....+{{C}_{n}}={{2}^{n}}-1\] \[[\because \,{{C}_{0}}={{\,}^{n}}{{C}_{0}}=1]\] Again, putting x = ?1, we get \[0={{C}_{0}}-{{C}_{1}}+{{C}_{2}}-{{C}_{3}}+....\] or \[{{C}_{0}}+{{C}_{2}}+{{C}_{4}}+....\]\[={{C}_{1}}+{{C}_{3}}+{{C}_{5}}+....\]i.e. \[A=B\] Also from (i), A + B = 2n or \[A={{2}^{n-1}}=B\] Hence, \[{{C}_{0}}+{{C}_{2}}+{{C}_{4}}+....={{2}^{n-1}}\].
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