[Actually $\displaystyle \lfloor5.7\rfloor$ is 5, not 7.]
The first three terms in that long sum will all be equal to 1 (because $\displaystyle \sqrt2$ and $\displaystyle \sqrt3$ both lie between 1 and 2). The next batch of terms, from $\displaystyle \lfloor\sqrt4\rfloor$ to $\displaystyle \lfloor\sqrt8\rfloor$ will all be equal to 2. Count how many of these terms there are. Then do the same for the batches of terms that are equal to 3, 4, 5, 6. Finally, the last two terms are equal to 7.
Now add all those numbers to get the answer.
Hello, rcs!
Opalg is absolutely correct!
Here's some trivia that shouldl help . . .
$\displaystyle \text{Evaluate: }\;[\sqrt{1}\,] + [\sqrt{2}\,] + [\sqrt{3}\,] + \hdots + [\sqrt{50}\,] $
. . $\displaystyle \text{where }[x]\text{ is the greatest integer function.}$
We note that squares differ by consecutive odd numbers.
. . $\displaystyle \begin{array}{cccccccccccc}
\text{Squares:} & 1 && 4 && 9 && 16 && 25 && \hdots \\
\text{Difference:} && 3 && 5 && 7 && 9 && \hdots \end{array}$
As Opalg explained:
. . $\displaystyle [\sqrt{1}\,],\;[\sqrt{2}\,],\;|\sqrt{3}\,]$ are all equal to 1.
. . $\displaystyle [\sqrt{4}\,],\;[\sqrt{5}\,],\;[\sqrt{6}\,],\;[\sqrt{7}\,],\;[\sqrt{8}\,]$ are all equal to 2.
. . $\displaystyle [\sqrt{9}\,]\;[\sqrt{10}\,],\; [\sqrt{11}\,]\;\hdots\;[\sqrt{15}\,]$ are all equal to 3.
. . and so on.
So we have:
. . $\displaystyle \begin{array}{ccc}
[\sqrt{1}\,] + [\sqrt{2}\,] + [\sqrt{3}\,] &=& 3(1) \\
[\sqrt{4}\,] + [\sqrt{5}\,] + \hdots + [\sqrt{8}\,] &=& 5(2) \\
[\sqrt{9}\,] + [\sqrt{10}\,] + \hdots + [\sqrt{15}\,] &=& 7(3) \\
[\sqrt{16}\,]+[\sqrt{17}\,] + \hdots + [\sqrt{24}\,] &=& 9(4) \\
[\sqrt{25}\,] + [\sqrt{26}\,] + \hdots [\sqrt{35}\,] &=& 11(5) \\
[\sqrt{36}\,] + [\sqrt{37}\,] + \hdots [\sqrt{48}\,] &=& 13(6) \\
[\sqrt{49}\,] + [\sqrt{50}\,] &=& 2(7) \\
& & --- \\
& & \text{Total?}
\end{array}$