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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Oct 2007 07:33:13 IST
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if [ r=1] [ n] tr =n(n+1)(n+2)(n+3)/12 ,find [r=1 ] [ infinity] 1/tr........ ?? the answer is 3/4!!!!!!!!
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SUCCESS IS HOW HIGH YOU BOUNCE AFTER HITTING THE BOTTOM.......
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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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SUCCESS IS HOW HIGH YOU BOUNCE AFTER HITTING THE BOTTOM.......
pls rate if my answers r helpfull....... |
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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
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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???
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SUCCESS IS HOW HIGH YOU BOUNCE AFTER HITTING THE BOTTOM.......
pls rate if my answers r helpfull....... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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!!!!!!!!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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,,,,,,,,
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Oct 2007 21:56:42 IST
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wow......in my method without partial fractions its sooooooo long
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SUCCESS IS HOW HIGH YOU BOUNCE AFTER HITTING THE BOTTOM.......
pls rate if my answers r helpfull....... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Oct 2007 21:59:27 IST
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it was an easy one
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you are in the competition to beat the competition
you work hard for a desired result |
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