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I have the sum of all these integer parts of different numbers $$ \lfloor 1\rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \dots + \lfloor \sqrt{15} \rfloor $$

I don't have any idea to solve this exercise.

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  • $\begingroup$ I wonder what the asymptotics are for $\lfloor \sqrt 1\rfloor + \lfloor \sqrt{2} \rfloor + \dots + \lfloor \sqrt{N} \rfloor$. $\endgroup$ – lhf Aug 5 '13 at 16:24
  • $\begingroup$ @lhf Note that: $$ \sum_{n=1}^N \lfloor \sqrt{n} \rfloor \approx \sum_{n=1}^N \sqrt{n} \approx \int_{1}^N \sqrt{x} ~dx = \Theta(N^{3/2}) $$ $\endgroup$ – Adriano Aug 5 '13 at 16:37
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More generally, for nonnegative integer $n$, $\lfloor \sqrt{x} \rfloor = n$ if $n \le \sqrt{x} < n+1$, i.e. $n^2 \le x < (n+1)^2$. Since $(n+1)^2 - n^2 = 2n+1$, there are $2n+1$ integers $x$ for which this is true. Thus $$\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2}\rfloor + \ldots + \lfloor \sqrt{(n+1)^2 - 1}\rfloor = \sum_{j=1}^n (2 j + 1) j = \dfrac{n(n+1)(4n+5)}{6}$$

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  • $\begingroup$ I don't understand something.Why did you write there (2j+1)j=n(n+1)(4n+5)/6?Can you give me more details? Where is that j coming from? $\endgroup$ – marinaaaa Aug 5 '13 at 19:04
  • $\begingroup$ The formulas for sums of $j$, $j^2$ etc. for $j$ from $1$ to $n$ are "well-known". $\endgroup$ – Robert Israel Aug 5 '13 at 22:18
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square root of 1 is 1

square root of 4 is 2

square root of 9 is 3

square root of 16 is 4

Logical :

$$\underbrace{1 \cdots 3}_{1}\ \underbrace{4 \cdots 8}_{2}\ \underbrace{9 \cdots 15}_{3}$$

Beauty is starting of each group is a perfect square . namely 1,4,9.

so square root of numbers from 1 to 3 it is 1

square root of numbers from 4 to 8 it is 2

square root of numbers from 9 to 15 it is 3

add all: 3 one's + 5 two's + 7 three's

so $ 3 + 10 + 21 $

= $34 $

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The greatest-integer function of $x$ finds the largest integer less than or equal to $x$. It is represented as $[x]$ or $\left\lfloor x \right\rfloor$. In the latter case, it is also called the floor function. Examples: $ [\pi] = 3, [-1] = -1, [6.9] = 6 $, etc. If you want sums of greatest-integer functions, you just want to try some values and try to make "blocks" of $[x]$ that give you the same result. In the case of $15$, it's quite simple, since the number isn't very big:

$ [\sqrt{1}] = [1] = 1 $

$ [\sqrt{2}] = 1 $

$ [\sqrt{3}] = 1 $

$ [\sqrt{4}] = 2 $

$ \cdots $

$ [\sqrt{15}] = 3 $

Also, this is not Calculus.

Beaten!

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