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 FunctionA. nB. n-1C. n+1D. n+2

Prove that for `n=1, 2, 3...``[(n+1)/2]+[(n+2)/4]+[(n+4)/8]+[(n+8)/16]

1.	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 FunctionA. nB. n-1C. n+1D. n+2
Answer» Correct Answer - A 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`.

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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 FunctionA. nB. n-1C. n+1D. n+2

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