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Asish Mahapatra
36 Points
7 years ago
[x] implies the greatest integer that is less than or equal to x.

Clearly [x] is 0 if x lies between 0 and 1 (exclusive) and [x] = 1 if x lies between 1 (inclusive) and 2 (exclusive)

In this case, the rth term is
$t_{r} = \left [ \frac{3}{4} + \frac{r-1}{100} \right ]$
This will be equal to zero till the value inside is atleast 1. Note that the last term is [3/4 + 99/100] = [1.74] = 1. So every term will equal 1, starting from the term where 3/4+ (r-1)/100 = 1
$\frac{3}{4} + \frac{r-1}{100} = 1$
$\implies \frac{r-1}{100} = \frac{1}{4}$
$\implies r-1 = 25$
$\boxed{\implies r = 26}$

Thus from r = 26, to r = 100, each term evaluates to 1.
So, the sum will be equal to: how many numbers are between 26 and 100 (both inclusive) = 100 – 26 + 1 = 75