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

 Jan 13, 2020
 #1
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Sumfor(n, 1, 100, floor(sqrt(n)) = 625

 Jan 13, 2020
 #2
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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

 

 

cool cool cool

 Jan 13, 2020

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