Let $\mathrm{f}\left(\mathrm{n}\right)=\left[\frac{1}{3}+\frac{3\mathrm{n}}{100}\right]$n, where [n] denotes the greatest integer less than or equal to n. Then $\sum _{\mathrm{n}=1}^{56}\mathrm{f}\left(\mathrm{n}\right)$ is equal to

$$\begin{gathered} \mathrm{f}\left(\mathrm{n}\right)=\left[\frac{1}{3}+\frac{3\mathrm{n}}{100}\right]\mathrm{n} \\ \sum _{\mathrm{n}=1}^{56} \mathrm{f}\left(\mathrm{x}\right)=0+0+0....+0+23+24....+55+2\times 56 \\ 22 \mathrm{terms} \\ =1399 \end{gathered}$$