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Algebra
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faadhil #
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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???
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16 Oct 2007 11:27:46 IST
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tr = [r=1 ]
[n ] tr - [r=1 ]
[n-1 ] tr
= n(n+1)(n+2) /3
Using partial fractions
[r=1 ]
[inf ] tr = 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
[n ] tr - [r=1 ]
[n-1 ] tr = n(n+1)(n+2) /3
Using partial fractions
[r=1 ]
[inf ] tr = 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
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)
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)
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!!!!!!!!!
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!!!!!!!!!


[ n] tr =n(n+1)(n+2)(n+3)/12 ,find [r=1 ]
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