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 the coefficients of odd powers of\[x\]. \[{{=}^{50}}{{C}_{1}}{{+}^{5}}{{C}_{3}}+...{{+}^{50}}{{C}_{49}}={{2}^{50-1}}\] \[={{2}^{49}}\]You need to login to perform this action.
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