Let [ ] denote the greatest integer function and f(x) = n 1n3 k=1n [ k23x ]. Then 12 j=1 f(j) is equal to .
Answer
2
Solution
[ k23x ] = k23x - \ k23x \, where 0 \ \ < 1
0 1n3 k=1n \ k23x \ < nn3 0
f(x) = n 13x 1n3 k=1n k2
= 13x n n(n + 1)(2n + 1)6n3 = 13x 13 = 13x+1
j=1 f(j) = 19 + 127 + = 191 - 13 = 16
12 j=1 f(j) = 12 16 = 2
Answer: 2