IIT JEE-AIEEE 2012, Entrance Notifications, Study Tips, Engineering Preparation, Practice Test Papers, Experts Advice

Titun

[ y ] denotes the greatest integer less than or equal to x

[ y ] = y - { y } where, {y} represents the fractional part of x.

Now, _{[n ]}

_{[infinity ]} ( [x] + [2x] + [3x] + ........+[nx] ) / n²

= _{[n ]}

_{[infinity ]} (x + 2x + 3x + ..........+ nx ) / n² - _{[n ]}

_{[infinity ]} ( {x} + {2x} + ............ + {nx} ) / n²

= _{[n ]}

_{[infinity ]} x . (1 + 2 + .........n) / n² - P

where, P = _{[n ]}

_{[infinity ]} ( {x} + {2x} + ............ + {nx} ) / n²

= _{[n ]}

_{[infinity ]} x . n (n + 1) / 2 . n² - P

= _{[n ]}

_{[infinity ]} x . (1 + 1/n) / 2 - P

= x / 2 - P

We know, that { y } for any real number y represents the fractional part of y
and 0

{ y } < 1

Therefore,
P = _{[n ]}

_{[infinity ]} ( {x} + {2x} + ............ + {nx} ) / n²

_{[n ]}

_{[infinity ]}( 1 + 1 + 1 +1 + 1.................. n times ) / n²

[ since, each term {x}, {2x}, {3x} ...........{nx} < 1 ]

i.e P

_{[n ]}

_{[infinity ]} n / n² = _{[n ]}

_{[infinity ]} 1 / n = 0

i.e P

0

But P can't be negative as each term of the limit is positive. {x}/n², {2x}/n²,.........{nx}/n²each of the terms are positive. So, the required limit has to be non-negative ( i.e either positive or zero )

Hence P must be zero as P can't be negative.

Hence,
_{[n ]}

_{[infinity ]} ( [x] + [2x] + [3x] + ........+[nx] ) / n²

= x / 2 - P = x / 2 - 0 = x / 2

Ans: x / 2

Cheers !!!!!!

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