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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Jan 2007 17:29:34 IST
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1+2[1+1/2007]2+3[1+1/2007]3+........to 2007 terms=
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2 Feb 2007 01:19:16 IST
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Add and subtract 1/2007 to the series and you will get
(1+1/2007)+2(1+1/2007)^2+3(1+1/2007)^3.......(upto 2700 terms)-(1/2700)
The first series is an arthematico geometric progression. Add it in the standandard way and you will get the answer. In case you can't solve this let me know and i will give you complete steps
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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4 Feb 2007 17:12:27 IST
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what is the standard formula to find sum in arithmetico geometric series
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4 Feb 2007 17:23:05 IST
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Spandana
The gen formula for Sn of AG series is (a/1-r) + (dr/(1-r)^2) -
dr^n/(1-r)^2 - (a+(n-1)d)r^n/1-r
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