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# Let $f(n) = \left[ {\dfrac{1}{2} + \dfrac{n}{{100}}} \right]$, where $\left[ . \right]$ denotes the greatest integer function. Then the value of $\sum\limits_{n = 1}^{151} {f(n)}$ isA) $101$B) $102$C) $104$D) $103$

Last updated date: 23rd Jul 2024
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Hint: Here we are given a function which is in greatest integer form. We have to find the summation of that function up to $151$. To do this first we find the value which is inside the $\left[ . \right]$ box. Then according to that value we find till which summation that value holds. After value changes repeat that same until we reach up to $151$. Then we add up all the values to get the value of the summation.

Complete step-by-step solution:
We know that till the value inside the box reaches $1$ the value of the function will be zero. That means,
$\dfrac{1}{2} + \dfrac{n}{{100}} = 1 \\ \Rightarrow \dfrac{n}{{100}} = 1 - \dfrac{1}{2} \\ \Rightarrow \dfrac{n}{{100}} = \dfrac{1}{2} \\ \Rightarrow n = 50$
Till $n = 49$, value of function is zero.
Now for the value of the function to be $2$, we get
$\dfrac{1}{2} + \dfrac{n}{{100}} = 2 \\ \Rightarrow \dfrac{n}{{100}} = 2 - \dfrac{1}{2} \\ \Rightarrow \dfrac{n}{{100}} = \dfrac{3}{2} \\ \Rightarrow n = 150$
Which means till $n = 149$, the value of the function will be $1$.
So the value of the function will be $2$ for $n = 150,151$.
Hence during the course of summation till $n = 151$, we get
$f(n) = 0$ for $49$ values
$f(n) = 1$ for $100$ values
$f(n) = 2$ for $2$ values
So, we get the summation as
$\sum\limits_{n = 1}^{151} {f(n)} = 0 \times 49 + 1 \times 100 + 2 \times 2 \\ \Rightarrow \sum\limits_{n = 1}^{151} {f(n)} = 100 + 4 \\ \Rightarrow \sum\limits_{n = 1}^{151} {f(n)} = 104$
Hence we get the answer as option C).

Note: We should take a note that whenever there is any function which is a greatest integer function, its value will always be the integer part of any number and the function will be constant till the very point where the value of the function changes to another integer. Means such a function will never give out a value with a fractional part.