Self proclaimed islands

Algebra Level 3

n = 1 100 n = ? \large \sum_{n=1}^{100} \lfloor \sqrt n \rfloor = \ ?


The answer is 625.

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1 solution

Alex Delhumeau
Jul 9, 2015

{ n = 1 if 1 n < 4 for a total of 3 cases n = 2 if 4 n < 9 for a total of 5 cases n = 3 if 9 n < 16 for a total of 7 cases \begin{cases} \lfloor \sqrt{n} \rfloor =1 &\mbox{if } 1 \leq n < 4 \text{ for a total of 3 cases} \\ \lfloor \sqrt{n} \rfloor = 2 & \mbox{if } 4\leq n < 9 \text{ for a total of 5 cases} \\ \lfloor \sqrt{n} \rfloor = 3 & \mbox{if } 9\leq n < 16 \text{ for a total of 7 cases} \end{cases}

...and so on up until n = 100 n=100 where have only 1 1 case and n = 10 \lfloor \sqrt{n} \rfloor = 10 .

Therefore, we can rewrite n = 1 100 n \sum_{n=1}^{100} \lfloor \sqrt n \rfloor as 10 + n = 1 9 ( 2 n + 1 ) n = 10 + 2 n = 1 9 n 2 + n = 1 9 n = 10 + 570 + 45 = 625 10+\sum_{n=1}^{9} (2n+1)n = \\ 10+2\sum_{n=1}^{9} n^2+\sum_{n=1}^{9} n = \\ 10+570+45 = \\ \boxed{625} .

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