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] New average \[=\,\,\frac{na+2+4+8+16+....+{{2}^{n}}}{n}\] |
| \[=\,\,\,\frac{na+2\left( \frac{{{2}^{n}}-1}{2-1} \right)}{n}=a+2\cdot \frac{{{2}^{n}}-1}{n}\] |
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