need algebra help

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Find the sum \(\displaystyle \sum_{n = 1}^{100} \lfloor \sqrt{n} \rfloor\)

Guest Jan 13, 2020

Guest · Answer 1 · 2020-01-13T06:23:38

Sumfor(n, 1, 100, floor(sqrt(n)) = 625

CPhill · Answer 2 · 2020-01-13T06:34:32

Note that

When n = 1 to n = 3....the sum = (1) *3 = 3

When n = 4 to n = 8, the sum = (2)* 5 = 10

When n = 9 to n =15, the sum = (3) * 7 = 21

So....generalizing this we have the sum

(1)*3 + (2)*5 + (3)*7 + ( 4)*9 +(5)*(11) + (6)*13 + (7)*15 + (8)*17 + (9)*19 + 10 = 625