Question

# If for a real number $$x,[x]$$ denotes the greatest integer less than or equal to $$x,$$ then for any $$n\in N$$, $$\displaystyle \left[ \frac { n+1 }{ 2 } \right] +\left[ \frac { n+2 }{ 4 } \right] +\left[ \frac { n+4 }{ 8 } \right] +\left[ \frac { n+8 }{ 16 } \right] +...=$$

A
n
B
n1
C
n+1
D
n+2

Solution

## The correct option is A $$n$$For any $$x\in R$$ we have$$\displaystyle \left[ x \right] =\left[ \frac { x }{ 2 } \right] +\left[ \frac { x+1 }{ 2 } \right]$$$$\displaystyle \Rightarrow \left[ n \right] =\left[ \frac { n+1 }{ 2 } \right] +\left[ \frac { n }{ 2 } \right]$$ ....... $$(i)$$$$\displaystyle \Rightarrow n=\left[ \frac { n+1 }{ 2 } \right] +\left[ \frac { n }{ 2 } \right]$$$$\displaystyle \Rightarrow n=\left[ \frac { n+1 }{ 2 } \right] +\left[ \frac { n }{ 4 } \right] +\left[ \frac { \frac { n }{ 2 } +1 }{ 2 } \right]$$      $$[$$using Eq, $$(i)]$$$$\displaystyle \Rightarrow n=\left[ \frac { n+1 }{ 2 } \right] +\left[ \frac { n+2 }{ 4 } \right] +\left[ \frac { n }{ 4 } \right]$$$$\displaystyle \Rightarrow n=\left[ \frac { n+1 }{ 2 } \right] +\left[ \frac { n+2 }{ 4 } \right] +\left[ \frac { n }{ 8 } \right] +\left[ \frac { \frac { n }{ 4 } +1 }{ 2 } \right]$$    $$[$$using Eq. $$(i)]$$$$\displaystyle \Rightarrow n=\left[ \frac { n+1 }{ 2 } \right] +\left[ \frac { n+2 }{ 4 } \right] +\left[ \frac { n+4 }{ 8 } \right] +\left[ \frac { n }{ 8 } \right]$$Continuing in this manner, we have$$\displaystyle \left[ \frac { n+1 }{ 2 } \right] +\left[ \frac { n+2 }{ 4 } \right] +\left[ \frac { n+4 }{ 8 } \right] +\left[ \frac { n+8 }{ 16 } \right] +...=n$$Mathematics

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