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Algebra

Cool goIITian

Joined:	12 Oct 2007
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16 Oct 2007 07:33:13 IST

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564

rise up to the challenge???

None

_{[ r=1]}

^{[ n]} t_{r ^{=n(n+1)(n+2)(n+3)/12 ,find}[r=1 ]}

^{[ infinity]} 1/tr........

the answer is 3/4!!!!!!!!

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Comments (8)

faadhil #

Cool goIITian

Joined: 12 Oct 2007

Posts: 49

16 Oct 2007 10:29:14 IST

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well?????? cant any`1 solve this shud i put up the solution???

Nadeem

Blazing goIITian

Joined: 25 Aug 2007

Posts: 521

16 Oct 2007 11:27:46 IST

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t_r= _{[r=1 ]}

^{[n ]} t_r - _{[r=1 ]}

^{[n-1 ]} t_r

= n(n+1)(n+2) /3

Using partial fractions
_{[r=1 ]}

^{[inf ]} t_r= 3/2 _{[r=1 ]}

^{[inf ]} 1/r + 3/2 _{[r=1 ]}

^{[inf ]} 1/r+2 - 3 _{[r=1 ]}

^{[inf ]} 1/r+1

=3/2( 1 + 1/2 +1/3 +1/4..) + 3/2( 1/3 +1/4 + 1/5......) - 3 ( 1/2+1/3.....)

= 3/2( 1+ 1/2) - 3( 1/2)

=3/4

faadhil #

Cool goIITian

Joined: 12 Oct 2007

Posts: 49

16 Oct 2007 12:32:05 IST

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oh hell,,,,,,,that was rocking man.....my sir gave me method thats 3 times as long........

what is partial fractions???

Nadeem

Blazing goIITian

Joined: 25 Aug 2007

Posts: 521

16 Oct 2007 13:08:41 IST

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you'll learn about partial fractions in detail when u learn about integration.

But its also useful in finding the sum of a series like the question above.

I'll just explain with an example.

Now u want to express 3/[n(n+1)(n+2) ] in terms of 1/n , 1/(n+1) , 1/(n+2)

So let 3/[n(n+1)(n+2) ]    =    A/n   + B/(n+1)   + C / (n+2)
We want to find A,B and C.

=> 3/[n(n+1)(n+2) ]    =    [A(n+1)(n+2) + B n(n+2) +C n(n+1)]/n(n+1)(n+2)

this is not an equation , it is true for all n.

So 3 =A(n+1)(n+2) + B n(n+2) +C n(n+1)
put n=-1
we get 3= B (-1)(1)
=> B=-3

put n=0
we get 3= A(2)
A=3/2

put n=-2
we get 3= C(-2)(-1)
C=3/2

So u get 3/n(n+1)(n+2)= (3/2) 1/n + (3/2) 1/(n+2) -     3 1/(n+1)

viswesh Nrayanan

Cool goIITian

Joined: 9 Oct 2007

Posts: 73

16 Oct 2007 15:46:31 IST

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I have a much simpler method for patial fraction instead of A B,C

In the fraction 3/[n(n+1)(n+2]
HEre subtitute n=0 in in all except n
u will get 3 / n2

now for secnd substitue -1( the zero of n+1) everywhere excpet in the n+1
u will get -3/n+1

now for the third take -2 (the zero of n+ 2)
Put in all except n+2
U will get (3/2) 1/(n+2)

put them together........THIs is very easy......IF U like RATE ME!!!!!!!!!

viswesh Nrayanan

Cool goIITian

Joined: 9 Oct 2007

Posts: 73

16 Oct 2007 15:48:03 IST

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HEY nadeemoidu (162)
I have made ur work much simpler.............IF u like this method RATE ME!!!!!!!!!!
almost the same as ur method,,,,,,,,

faadhil #

Cool goIITian

Joined: 12 Oct 2007

Posts: 49

16 Oct 2007 21:56:42 IST

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wow......in my method without partial fractions its sooooooo long

Ambareesh Rishi

Blazing goIITian

Joined: 4 May 2007

Posts: 499

16 Oct 2007 21:59:27 IST

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it was an easy one

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