Question

Let\[f\left( n \right) = \left[ {\dfrac{1}{3} + \dfrac{{3n}}{{100}}} \right]n\] , where [n] denotes the greatest integer less than or equal to n. Then \[\sum\nolimits_{n = 1}^{56} {f(n)} \] is equal to
A. 56
B. 689
C. 1287
D. 1399

Answer 1

Hint: Here, find f (1), f (2), …, upto some terms and see the pattern. Find the values and if they form a sequence and then use formula.

Complete step-by-step answer:
We have \[f\left( n \right) = \left[ {\dfrac{1}{3} + \dfrac{{3n}}{{100}}} \right]n\]
Putting n = 1, 2, 3 …
\[\Rightarrow f\left( 1 \right) = \left[ {\dfrac{1}{3} + \dfrac{3}{{100}}} \right] \times 1 = \left[ {0.3 + 0.03} \right] \times 1 = \left[ {0.33} \right] \times 1 = 0 \times 1 = 0\]
\[\Rightarrow f\left( 2 \right) = \left[ {\dfrac{1}{3} + \dfrac{{3 \times 2}}{{100}}} \right] \times 2 = \left[ {0.3 + 0.06} \right] \times 2 = \left[ {0.36} \right] \times 2 = 0 \times 2 = 0\]
\[\Rightarrow f\left( 3 \right) = \left[ {\dfrac{1}{3} + \dfrac{{3 \times 3}}{{100}}} \right] \times 2 = \left[ {0.3 + 0.09} \right] \times 2 = \left[ {0.39} \right] \times 3 = 0 \times 3 = 0\]
.
.
.
\[\Rightarrow f\left( {22} \right) = \left[ {\dfrac{1}{3} + \dfrac{{3 \times 22}}{{100}}} \right] \times 22 = \left[ {0.3 + 0.66} \right] \times 22 = \left[ {0.96} \right] \times 22 = 0 \times 22 = 0\]
We can see that f (1) to f (22) = 0
Now,
\[\Rightarrow f\left( {23} \right) = \left[ {\dfrac{1}{3} + \dfrac{{3 \times 23}}{{100}}} \right] \times 23 = \left[ {\dfrac{{307}}{{300}}} \right] \times 23 = \left[ {1.02} \right] \times 23 = 1 \times 23 = 23\]
\[f\left( {24} \right) = \left[ {\dfrac{1}{3} + \dfrac{{3 \times 24}}{{100}}} \right] \times 24 = \left[ {0.3 + 0.72} \right] \times 24 = \left[ {1.02} \right] \times 24 = 1 \times 24 = 24\]
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     .
     .
\[\Rightarrow f\left( {56} \right) = \left[ {\dfrac{1}{3} + \dfrac{{3 \times 56}}{{100}}} \right] \times 56 = \left[ {\dfrac{{604}}{{300}}} \right] \times 56 = \left[ {2.01} \right] \times 56 = 2 \times 56 = 112\]
f (23) + f (24) +...+ f (56) = 23 + 24 + ... + 55 + 2 × 56
Here, 23 + 24 + ... + 55 forms an AP with first term = 23, common difference = 1 and last term = 55.
Let nth term is 55
55 = 23 + (n – 1) × 1 = 22 + n ⇒ n = 33
Sum = $ \dfrac{{33}}{2}\left( {23 + 55} \right) = 33 \times 39 $
Now, 23 + 24 + ... + 55 + 112 = 33 × 39 + 112 = 1287 + 112 = 1399
\[\sum\nolimits_{n = 1}^{56} {f(n)} \]= 1399
So, the correct answer is “Option D”.

Note: In these types of questions, find the value keeping in mind the greatest integer function. Do not find value at each value because the solution becomes more complicated. Try to make a pattern in results obtained by putting each value.
The Greatest Integer Function is denoted by f(x) = [x].
For all real numbers, x, the greatest integer function gives the largest integer less than or equal to x. It rounds down a real number to the nearest integer.
For example: [1.2] = 1, [1.7] = 1, [3.2] = 3, [4.4] = 4, [– 3] = – 2, [– 1.8] = – 2, [– 2.3] = – 3, [– 5.4] = – 6