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

Question

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

Smart Achievers · Accepted Answer

(825)

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

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

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

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

Solution

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

Want Us to Email you About Special Offers & Updates?

Solution

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

Ans. (825) ∑r=1120[r]=∑r=131+∑r=482+∑r=9153+∑r=16254+…+∑r=81999+∑r=10012010 =3(1)+5(2)+7(3)+9(4)+…+21(10) =∑r=110(2r+1)r=2×10×11×216+10×112 =10×112×15=825

Want Us to Email you About Special Offers & Updates?

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