A) \[{{2}^{n}}\]
B) \[{{2}^{n}}-1\]
C) 0
D) \[{{2}^{n-1}}\]
Correct Answer: C
Solution :
We know that \[{{(1+x)}^{n}}={{\,}^{n}}{{C}_{0}}+{{\,}^{n}}{{C}_{1}}x+{{\,}^{n}}{{C}_{2}}{{x}^{2}}+....+{{\,}^{n}}{{C}_{n}}{{x}^{n}}\] Putting x = ?1, we get \[{{(1-1)}^{n}}={{\,}^{n}}{{C}_{0}}-{{\,}^{n}}{{C}_{1}}+{{\,}^{n}}{{C}_{2}}-.....{{(-1)}^{n\,\,n}}{{C}_{n}}\] Therefore \[{{C}_{0}}-{{C}_{1}}+{{C}_{_{2}}}-{{C}_{3}}+....(-1){{\,}^{n}}{{C}_{n}}=0\]
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