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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
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B
n1
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C
n+1
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D
n+2
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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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