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 $\sum _{n=1}^{56}{\Delta }_{r}f\left(n\right)$ is equal to :

• A. 56
• B. 689
• C. 1287
• D. 1399

Answer 1

Answer: (D)

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