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Let $f\left( n \right) = \left[ {\frac{1}{3} + \frac{{3n}}{{100}}} \right]n$ , where $[n]$ denotes the greatest integer less than or equal to $n$. Then $\sum\limits_{n = 1}^{56} {f\left( n \right)} $ is equal to
$56$
$689$
$1287$
$1399$
Solution
Let $f\left( n \right) = \left[ {\frac{1}{3} + \frac{{3n}}{{100}}} \right]n$
where $\left[ n \right]$ is greatest integer functon,
$ = \left[ {0.33 + \frac{{3n}}{{100}}} \right]n$
For $n = 1,2,….,22,$ we get $f\left( n \right) = 0$
and for $n = 23,24,….,55,$ we get $f\left( n \right) = 1$
For $n = 56,f\left( n \right) = 2$
So, $\sum\limits_{n = 1}^{56} {f\left( n \right) = 1\left( {23} \right) + 1\left( {24} \right) + … + 1\left( {55} \right)} + 2\left( {56} \right)$
$ = \left( {23 + 24 + ….. + 55} \right) + 112$
$ = \frac{{33}}{2}\left[ {46 + 32} \right] + 112$
$ = \frac{{33}}{2}\left( {78} \right) + 112 = 1399$.
