A) 56
B) 689
C) 1287
D) 1399
Correct Answer: D
Solution :
Let\[f(n)=\left[ \frac{1}{3}+\frac{3n}{100} \right]n\] where [n] is greatest integer function, \[=\left[ 0.33+\frac{3n}{100} \right]n\] For n = 1, 2, ..., 22, we get f (n) = 0 and for n = 23, 24, ..., 55, we get f (n) = 1 For n = 56, f (n) = 2 So, \[\sum\limits_{n=1}^{56}{f(n)=1}(23)+1(24)+...+1(55)+2(56)\] \[=(23+24+...+55)+112\] \[=\frac{33}{2}[46+32]+112\] \[=\frac{33}{2}(78)+112=1399.\]You need to login to perform this action.
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