Let f(n) = left[ frac{1}{3} + frac{3n}{100} r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(n) = \left[ \frac{1}{3} + \frac{3n}{100} \right]n$. For $1 \le n \le 22$,$\frac{1}{3} + \frac{3n}{100} < \frac{1}{3} + \frac{66}{100} = 0

Vedclass · Accepted Answer

$1399$

Vedclass · Answer

(D) Given $f(n) = \left[ \frac{1}{3} + \frac{3n}{100} ight]n$. For $1 \le n \le 22$,$\frac{1}{3} + \frac{3n}{100} For $23 \le n \le 55$,$1 \le \frac{1}{3} + \frac{3n}{100} For $n = 56$,$\frac{1}{3} + \frac{3(56)}{100} = 0.333 + 1.68 = 2.013$. Thus,$\left[ \frac{1}{3} + \frac{3(56)}{100} ight] = 2$,so $f(56) = 2 	imes 56 = 112$. The sum is $\sum_{n=1}^{56} f(n) = \sum_{n=1}^{22} 0 + \sum_{n=23}^{55} n + f(56)$. $= 0 + \frac{(55-23+1)}{2}(23+55) + 112 = \frac{33}{2}(78) + 112 = 33 	imes 39 + 112 = 1287 + 112 = 1399$.

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

Similar Questions

The sum to $n$ terms of the series $2^2 + 4^2 + 6^2 + \dots$ is

Find the sum of the following series up to $n$ terms:
$0.6 + 0.66 + 0.666 + \dots$

$\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j 1 } } = \dots$

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$ to $n$ terms $= n(n+1) f(n)$,then $f(2) =$

The sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module