Let denote the greatest integer . Then [ 1 ] + [ 2 ] + [ 3 ] + . . . + [ 120 ] is equal to __________ . [2023]

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Solution

Q.
Let $[α]$ denote the greatest integer $\leq α$ . Then $[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + . . . + [\sqrt{120}]$ is equal to __________ . [2023]

Ans.
(825)

$\sum_{r = 1}^{120} [\sqrt{r}] = \sum_{r = 1}^{3} 1 + \sum_{r = 4}^{8} 2 + \sum_{r = 9}^{15} 3 + \sum_{r = 16}^{25} 4 + \dots + \sum_{r = 81}^{99} 9 + \sum_{r = 100}^{120} 10$

$= 3 (1) + 5 (2) + 7 (3) + 9 (4) + \dots + 21 (10)$

$= \sum_{r = 1}^{10} (2 r + 1) r = 2 \times \frac{10 \times 11 \times 21}{6} + \frac{10 \times 11}{2}$

$= \frac{10 \times 11}{2} \times 15 = 825$

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