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FUNCTION-JEE (Question)

ANKIT PANDE said – Sat, 27 Jun 2009 15:11:26 -0000 ( Edit Link )

a function f is defined for all positive integers and satisfies f(1)=2005 and f(1)+f(2)—-——f(n)=n^2f(n) for all n>1.find value of f(2004).please explain the methodology.

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Author: ANKIT PANDE; Authority 80

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mahesh25 said – Fri, 03 Jul 2009 10:13:54 -0000 ( Flag Edit Mark / Unmark Best Link )

given that f(1)=2005

f(1) + f(2) + f(3) +...............+f(n)=n^2 .f(n) for all n>1 .................(1)

This means for n=1 f(1)=2005 and for the above values > 1 i.e n=2,3,4….... the values can be found through the equation 1

to find f(2) put n= 2 in equation 1 we get

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Author: mahesh25; Authority 235

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Author: Sureshbala; Authority 787

Sureshbala said – Mon, 06 Jul 2009 08:02:33 -0000 ( Flag Edit Mark / Unmark Best Link )

$f(1) + f(2) + ..............+ f(n-1) + f(n) = n^2 \times f(n)$

$\Rightarrow f(1) + f(2) + ........ + f(n-1) = n^2 \times f(n) - f(n)$

$\Rightarrow f(1) + f(2) + ........ + f(n-1) = f(n)[(n^2 - 1)] .......(1)$

By the given definition , we have

$\Rightarrow f(1) + f(2) + ........ + f(n-1) = (n-1)^2 \times f(n-1) .......(2)$

From (1) and (2), we have

$f(n)[(n^2 - 1)] = f(n-1)\times (n-1)^2$

$\Rightarrow \frac{f(n)}{f(n-1)} = \frac{n-1}{n+1}$

Now, we have

$\frac{f(2)}{f(1)} \times \frac{f(3)}{f(2)} \times \frac{f(4)}{f(3)} \times \frac{f(5)}{f(4)} \times..$ $....\times \frac{f(2003)}{f(2002)} \times\frac{f(2004)}{f(2003)}$

$= \frac{1}{3} \times \frac{2}{4} \times \frac{3}{5} \times \frac{4}{6} \times ..$ $..... \times \frac{2002}{2004} \times\frac{2003}{2005}$

$\Rightarrow \frac{f(2004)}{f(1)} = \frac{2}{2004 \times 2005}$

$\Rightarrow \frac{f(2004)}{2005} = \frac{2}{2004 \times 2005}$

$\Rightarrow f(2004) = \frac{1}{1002}$

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