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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Jun 2008 10:50:21 IST
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[url=http://www.signaturebar.com/][img]
B DESIGN
IIT GUWAHATI. |
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18 Jun 2008 12:06:41 IST
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let S=6/5+27/25+128+125................
S=1+1/5+1+2/5+1+3/125...........
S=n+1/5+2/25+3/125..............n/5^n
S/5=n/5+1/25+2/125..............(n-1)/5^n+n/5^(n+1)
subtracting
4S/5=4n/5+1/5+1/25................1/5^5+n/5^(n+1)
4S/5=4n/5+(1-(1/5)^n)/4+n/5^(n+1)
cheers!
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