A) 0
B) \[{{2}^{49}}\]
C) \[{{2}^{50}}\]
D) \[{{2}^{51}}\]
Correct Answer: B
Solution :
We have, \[{{(1+x)}^{50}}=\sum\limits_{r=0}^{50}{{}^{50}{{C}_{r}}{{x}^{r}}}\]. Therefore, sum of coefficients of odd power of x = \[{}^{50}{{C}_{1}}+{}^{50}{{C}_{3}}+...+{}^{50}{{C}_{49}}\] = \[\frac{1}{2}[{}^{50}{{C}_{0}}+{}^{50}{{C}_{1}}+...+{}^{50}{{C}_{50}}]\,\,=\,\,\frac{1}{2}[{{2}^{50}}]={{2}^{49}}\].
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