Question

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

• A. $56$
• B. $689$
• C. $1287$
• D. $1399$

Answer 1

The correct option is D $1399$

$f\left(x\right)=[\frac{1}{3}+\frac{3x}{100}]x$
$f\left(1\right)$ to $f\left(22\right)=0$
$f\left(23\right)=[\frac{1}{3}+\frac{69}{100}]23$
$f\left(24\right)=[\frac{1}{3}+\frac{72}{100}]24$
$f\left(23\right)+f\left(24\right).....f\left(56\right)=23+24+.....+55+2\times 56$
$=33\times 39+112$
$=1399$ .