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[√1]+[√2]+[√3]+...+[√100]
1- 3 times (for √1, √2, √3)
2 - 5 times ( for √4,√5,√6,√7,√8)
.
.
.
10 - 1 time (for √100)


so sum : { 1*3 + 2*5 + 3*7 +... 9 terms } + 1* 10
= > Sum{n(3+(n-1)*2) + 10 = Sum{n(2n + 1)}|9 terms + 10 => sum(2n^2 + n)|9 terms + 10
=> 570 + 45 + 10 = 625

Ans should be
B
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Bunuel
If [X] denotes the greatest integer less than or equal to x, what is the value of \([√1] + [√2] + [√3] + ... + [√100]\) ?

A. 615
B. 625
C. 505
D. 5050
E. 5125


Let’s determine the number of terms whose values are 1, 2, 3, etc.

Value of 1: [√1], [√2], [√3] → 3 terms

Value of 2: [√4], [√5], …, [√8] → 5 terms

Value of 3: [√9], [√10], …, [√15] → 7 terms

Without continuing the list, we can see that there are 9 terms with the value of 4, 11 with 5, 13 with 6, 15 with 7, 17 with 8 and 19 with 9. Finally, there is 1 more term, the last term, with the value of 10.

Therefore, the sum of all 100 terms is:

1 x 3 + 2 x 5 + 3 x 7 + 4 x 9 + 5 x 11 + 6 x 13 + 7 x 15 + 8 x 17 + 9 x 19 + 10 x 1

3 + 10 + 21 + 36 + 55 + 78 + 105 + 136 + 171 + 10 = 625

Answer: B
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What is the basic knowledge needed to solve this one? I am not aware why we would be grouping the numbers. Thanks!
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What is the basic knowledge needed to solve this one? I am not aware why we would be grouping the numbers. Thanks!
Because [X] denotes the greatest integer less than or equal to x, we group the numbers that come before any perfect square, for example \(\sqrt{4} = 2\), every square root before \(\sqrt{4}\) will give 1.something, and since it is inside [], as per definition of [X] it will become integer 1. Similarly, anything from \(\sqrt{5}\) to \(\sqrt{8}\) will become 2.
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