Let fleft( n right) = left[ {frac{1}{3} + frac{{3n}}{{100}}} right]n , where [n] denotes the&n

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1.Relation and Function

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

A

$56$

B

$689$

C

$1287$

D

$1399$

(JEE MAIN-2014)

Solution

Let $f\left( n \right) = \left[ {\frac{1}{3} + \frac{{3n}}{{100}}} \right]n$

where $\left[ n \right]$ is greatest integer functon,

$ = \left[ {0.33 + \frac{{3n}}{{100}}} \right]n$

For $n = 1,2,….,22,$ we get $f\left( n \right) = 0$

and for $n = 23,24,….,55,$ we get $f\left( n \right) = 1$

For $n = 56,f\left( n \right) = 2$

So, $\sum\limits_{n = 1}^{56} {f\left( n \right) = 1\left( {23} \right) + 1\left( {24} \right) + … + 1\left( {55} \right)} + 2\left( {56} \right)$

$ = \left( {23 + 24 + ….. + 55} \right) + 112$

$ = \frac{{33}}{2}\left[ {46 + 32} \right] + 112$

$ = \frac{{33}}{2}\left( {78} \right) + 112 = 1399$.

Standard 12

Mathematics

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