Prove that for `n=1, 2, 3...` `[(n+1)/2]+[(n+2)/4]+[(n+4)/8]+[(n+8)/16]+...=n` where `[x]` represents Greatest Integer Function

Options

A. n
B. n-1
C. n+1
D. n+2

Answer

A

Explanation

For any ` x in R `, we have
`[x]=[(x)/(2)]+[(x+1)/(2)]" "`….(i)
`implies [n]=[(n+1)/(2)]+[(n)/(2)]`
`implies n=[(n+1)/(2)]+[(n)/(2)]`
`implies n=[(n+1)/(2)]+[(n)/(2)]+[((n)/(2)+1)/(2)]" " ` [ using (i)]
`implies n=[(n+1)/(2)]+[(n+2)/(4)+[(n)/(4)]`
`implies n=[(n+1)/(2)]+[(n+2)/(4)]+[(n)/(8)]+[((n)/(4)+1)/(2)]" "`[using (i)]
`implies n[(n+1)/(2)]+[(n+2)/(4)]+[(n+4)/(8)]+[(n)/(8)]`
Continuing in this manner , we have
`[(n+1)/(2)]+[(n+2)/(4)]+[(n+4)/(8)]+[(n+8)/(16)]+....=n`.