Sum of integer parts of different numbers 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.
 A: 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
$$\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2}\rfloor + \ldots + \lfloor \sqrt{(n+1)^2 - 1}\rfloor = \sum_{j=1}^n (2 j + 1) j = \dfrac{n(n+1)(4n+5)}{6}$$ 
A: 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 $
A: 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! 
