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![If [n] is defined as the greatest integer less than or equal to n](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If [n] is defined as the greatest integer less than or equal to n"){kind=icon}
Updated on: 17 May 2021, 14:04
Originally posted by sujoykrdatta on 14 May 2021, 12:18.
Last edited by sujoykrdatta on 17 May 2021, 14:04, edited 1 time in total.
Last edited by sujoykrdatta on 17 May 2021, 14:04, edited 1 time in total.
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C
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Question Stats:
60% (02:16) correct
40%
(02:18)
wrong
based on 229
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If [n] is defined as the greatest integer less than or equal to n, for x = 1, 2, 3, … n, what is the value of:
\([\frac{(x+1)}{2}]\) + \([\frac{(x+2)}{4}]\) + \([\frac{(x+4)}{8}]\) + ... \(n\) terms?
A) x–2
B) x-1
C) x
D) x+2
E) x+3
\([\frac{(x+1)}{2}]\) + \([\frac{(x+2)}{4}]\) + \([\frac{(x+4)}{8}]\) + ... \(n\) terms?
A) x–2
B) x-1
C) x
D) x+2
E) x+3
14 May 2021, 20:42
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sujoykrdatta
Easiest way would be to substitute x as some values.
x=1.... \([\frac{(x+1)}{2}] + [\frac{(x+2)}{4}] + [\frac{(x+4)}{8}] + ... n \)= \([\frac{(1+1)}{2}] + [\frac{(1+2)}{4}] + [\frac{(1+4)}{8}] + ... n \)= \([\frac{(2)}{2}] + [\frac{(3)}{4}] + [\frac{(5)}{8}] + ... n \)=1+0+0=1=x
Only C fits in
x=2.... \([\frac{(x+1)}{2}] + [\frac{(x+2)}{4}] + [\frac{(x+4)}{8}] + ... n \)= \([\frac{(2+1)}{2}] + [\frac{(2+2)}{4}] + [\frac{(2+4)}{8}] + ... n \)= \([\frac{(3)}{2}] + [\frac{(4)}{4}] + [\frac{(6)}{8}] + ... n \)=1+1+0=2=x
Only C fits in
C
15 May 2021, 16:13
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sujoykrdatta
We can plug in x = 1, which has only one term and that one term gives [1] = 1. Then the answer would be 1, in terms of x that would be x so we choose C.
Ans: C
