Mathematics - Sequence & Series Question with Solution | TestHub

Given $f\left(n\right)=\left[\frac{1}{3}+\frac{3n}{\mathrm{100}}\right]n=\left[\frac{9n+\mathrm{100}}{\mathrm{300}}\right]n$

If $\frac{9n+100}{300}<1$

$\Rightarrow 9n+100<300$

$\Rightarrow n<\frac{\mathrm{200}}{9}$

$\Rightarrow n=22$

We know that for a number $0⩽x<1, \left[x\right]=0$

$\Rightarrow f\left(n\right)=0, \forall 1⩽n⩽22$

Again, if $\frac{9n+100}{300}<2$

$\Rightarrow 9n+100<600$

$\Rightarrow n=\frac{\mathrm{500}}{9}$

$\Rightarrow n=\mathrm{55}$

We know that for a number $1⩽x<2, \left[x\right]=1$

$\Rightarrow f\left(n\right)=1, 23⩽n⩽55$

And, $f\left(n\right)=2$ if $n=56$

Thus, $\sum _{n=1}^{\mathrm{56}}f\left(n\right)=\sum _{n=1}^{22}0\times n+\sum _{n=23}^{55}1\times n+2\times 56$

           $=0+\left(\mathrm{23}+\mathrm{24}+...+\mathrm{55}\right)+\mathrm{56}\times 2$

          $=\frac{33}{2}\left(23+55\right)+112$

    $=33\times 39+112$

    $=1399.$