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JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • question_answer
    Let \[f(n)=\left[ \frac{1}{3}+\frac{3n}{100} \right]n,\] 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:     JEE Main Online Paper (Held On 19 April 2016)

    A) 56                                          

    B) 689

    C) 1287                      

    D) 1399

    Correct Answer: D

    Solution :

    Let\[f(n)=\left[ \frac{1}{3}+\frac{3n}{100} \right]n\]  where [n] is greatest integer function, \[=\left[ 0.33+\frac{3n}{100} \right]n\] For n = 1, 2, ..., 22, we get f (n) = 0 and for n = 23, 24, ..., 55, we get f (n) = 1 For n = 56, f (n) = 2 So, \[\sum\limits_{n=1}^{56}{f(n)=1}(23)+1(24)+...+1(55)+2(56)\] \[=(23+24+...+55)+112\] \[=\frac{33}{2}[46+32]+112\] \[=\frac{33}{2}(78)+112=1399.\]


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