Question

Let f(n) = $[\frac{1}{2}+\frac{(n-1)}{50}]\left(n^2\right)$, where [n] denotes the greatest integer less than or equal to n, then ${\sum }_{n=1}^{100}f\left(n\right)$ is equal to;

• A. 295425
• B. 395425
• C. 295400
• D. 295475

Answer 1

The correct option is A 295425

$Givenf\left(n\right)=[\frac{1}{2}+\frac{(n-1)}{50}]\left(n^2\right)$
$f\left(1\right)=f\left(2\right)=\cdots =f\left(50\right)=0$
$f\left(51\right)={51}^2,f\left(52\right)={52}^2,\cdots ,f\left(100\right)={100}^2$
$\therefore {\sum }_{n=1}^{100}f\left(n\right)={51}^2+{52}^2+{53}^2+\cdots +{100}^2$
$=\sum {100}^2-\sum {50}^2$
$=\frac{100\times 101\times 201}{6}-\frac{50\times 51\times 101}{6}$
$=295425$