A) \[n-1\]
B) n
C) \[n+2\]
D) 2n
Correct Answer: B
Solution :
We have \[n=\left[ \frac{n}{2} \right]+\left[ \frac{n+1}{2} \right]\]or |
\[n=\left[ \frac{n+1}{2} \right]+\left[ \frac{n}{4} \right]+\left[ \frac{n+2}{4} \right]\] |
(Using (1) again and again) |
\[=\left[ \frac{n+1}{2} \right]+\left[ \frac{n+2}{4} \right]+\left[ \frac{n+4}{8} \right]+.....\] |
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