Question

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

• A. 56
• B. 689
• C. 1287
• D. 1399

Answer 1

The Correct Option is D

$\displaystyle\sum_{n=1}^{56} f(x)=\left[\frac{1}{3}+\frac{3 \times 1}{100}\right] \times 1+\ldots . .+\left[\frac{1}{3}+\frac{3 \times 22}{100}\right] \times 22 $
$+\left[\frac{1}{3}+\frac{3 \times 23}{100}\right] \times 23$ ........ + .........
${\left[\frac{1}{3}+\frac{3 \times 55}{100}\right] \times 55+\left[\frac{1}{3}+\frac{3 \times 56}{100}\right] \times 56} $
$=0+\ldots \ldots . .+0+23+24+\ldots \ldots .+55+2 \times 56 $
$=\frac{55(56)}{2}-\frac{22(23)}{2}+112 $
$=11(5 \times 28-23)+112$
$=11 \times 117+112 $
$=1287+112 $
$=1399$