A) \[a+\frac{{{2}^{n+1}}}{n}\]
B) \[a+2\frac{{{2}^{n}}-1}{n}\]
C) \[a+\frac{{{2}^{n}}-1}{n}\]
D) \[a+\frac{{{2}^{n\,+\,1}}-1}{n}\]
Correct Answer: B
Solution :
| [b] Total sum of n number \[=na\] According to the question, New average of numbers \[=\frac{na+(2+4+8+16+...)}{n}\] \[=a+\frac{2\,\,({{2}^{n}}-1)}{\frac{2-1}{n}}\] \[=a+\frac{2\,\,({{2}^{n}}-1)}{n}\] |
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