Question

Find the value of \[\left[ {\dfrac{3}{4}} \right] + \left[ {\dfrac{3}{4} + \dfrac{1}{{100}}} \right] + \left[ {\dfrac{3}{4} + \dfrac{2}{{100}}} \right] + ..... + \left[ {\dfrac{3}{4} + \dfrac{{99}}{{100}}} \right]\]

Answer 1

Hint: In the given question we need to find the value of the sum of the series.
For this case we will be using the condition such that:
\[\left[ x \right] = 0\] if \[x < 1\]
\[[x] = 1\] if \[2 > x > 1\]
Where the decimal values are made to be round of with 0 or 1 value.

Complete step-by-step answer:
In this problem,
Given: \[\left[ {\dfrac{3}{4}} \right] + \left[ {\dfrac{3}{4} + \dfrac{1}{{100}}} \right] + \left[ {\dfrac{3}{4} + \dfrac{2}{{100}}} \right] + ..... + \left[ {\dfrac{3}{4} + \dfrac{{99}}{{100}}} \right]\]
The value of \[\left[ {\dfrac{3}{4}} \right]\] = [0.75] = 0,
\[\left[ {\dfrac{3}{4} + \dfrac{1}{{100}}} \right]\]= [0.76] = 0
Until this the values are round off and we got the value as 0.
By the conditioned mentioned above,
\[\left[ x \right] = 0\] if \[x < 1\]
\[[x] = 1\]if \[2 > x > 1\]
For rounding up off decimal values to find the whole number using the sum of series.
\[\left[ {\dfrac{3}{4} + \dfrac{{25}}{{100}}} \right]\]= 1
i.e. Applying in the general term to find the first time of value 1.
\[\sum\limits_{r = 0}^{99} {\left[ {\dfrac{3}{4} + \dfrac{r}{{100}}} \right]} \]=1
Equating the value, to find the value of r.
\[\dfrac{r}{{100}} + \dfrac{3}{4} = 1\]
\[\dfrac{r}{{100}} = 1 - \dfrac{3}{4}\]
\[\dfrac{r}{{100}} = \dfrac{{4 - 3}}{4}\]
\[
  \dfrac{r}{{100}} = \dfrac{1}{4} \\
  r = \dfrac{1}{4} \times 100 \\
  r = 25 \\
 \]
On Calculating we find the value of r as 25, hence the first term with the value of 1 occurs from the 25th term.
Hence, with the following
\[\left[ {\dfrac{3}{4} + \dfrac{{26}}{{100}}} \right]\]= [1.01] = 1
\[\left[ {\dfrac{3}{4} + \dfrac{{99}}{{100}}} \right]\] = [1.74] = 1
(here the value becomes 1 due to the condition which we mentioned above)
Thus, \[\left[ {\dfrac{3}{4} + \dfrac{{26}}{{100}}} \right] + ..... + \left[ {\dfrac{3}{4} + \dfrac{{99}}{{100}}} \right]\] = 1+1+………. +1 = 74 (74 times repeats)
And with this combination \[\left[ {\dfrac{3}{4} + \dfrac{{25}}{{100}}} \right]\] = 1
Totally 75 times.
So, we have to solve \[\left[ {\dfrac{3}{4}} \right] + \left[ {\dfrac{3}{4} + \dfrac{1}{{100}}} \right] + \left[ {\dfrac{3}{4} + \dfrac{2}{{100}}} \right] + ..... + \left[ {\dfrac{3}{4} + \dfrac{{99}}{{100}}} \right]\]
     \[\begin{array}{*{20}{l}}
  { = \left[ {0.75} \right] + \left[ {0.76} \right] + \left[ {0.77} \right] + \ldots \ldots \ldots . + \left[ {0.99} \right] + \left[ 1 \right] + \left[ {1.01} \right] + \ldots \ldots . + \left[ {1.74} \right]} \\
  { = 0 + 0 + 0 + \ldots \ldots \ldots . + 0 + 1 + 1 + ....... + {\text{ }}1{\text{ }}\left( {75{\text{ }}times} \right)} \\
  {\; = 75.}
\end{array}\]
Therefore, \[\left[ {\dfrac{3}{4}} \right] + \left[ {\dfrac{3}{4} + \dfrac{1}{{100}}} \right] + \left[ {\dfrac{3}{4} + \dfrac{2}{{100}}} \right] + ..... + \left[ {\dfrac{3}{4} + \dfrac{{99}}{{100}}} \right]\] = 75

Note: For this kind of sum, we need to focus on the general term part rather than formula addicting level.
Majorly, after converting to general be consecutive on the term occurrence which is to be made to count to get the correct count without any error of calculation.