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Differential Calculus

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15 May 2010 10:06:29 IST

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evalute 1/n^100 lim as r tends to infinity find summation ( r goes from 1 to n ) ( r^99)

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evalute 1/n^100 lim as r tends to infinity find summation ( r goes from 1 to n ) ( r^99)

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Deepak Aggarwal

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Joined: 9 May 2009

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16 May 2010 20:50:12 IST

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Ans is 1/100..........

Akhil

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Joined: 26 Mar 2010

Posts: 112

16 May 2010 22:46:47 IST

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HOW TO WE GET THAT RESULT..

COULD YOU PLS PROVE THAT RESULT

Filokalk d' nix

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17 May 2010 13:36:41 IST

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another way is to use integration for limit of sum

S= lim (n->infinity) summation (1 to n){ (r/n)^99 * 1/n}

= integral(0 to 1) x^99 dx

=[(x^100)/100]..........( use limits 0&1

=1/100

vijay kharya

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23 May 2010 09:48:56 IST

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@ filokalk shouldn't the limits used be from 1 to n because the summation is from 1 to n.

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