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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 Apr 2008 14:59:09 IST
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Answer is 4e - 3 .....
I think i solved it before, however
look dude
S = 3 + 5 + 9 + 15 + 23 + ....... tn
S = 3 + 5 + 9 + 15 + ..............tn
Now subtract the two
so 0 = 3 + 2 + 4 + 6 + 8 + ......- tn
or tn = 3 + (2 + 4 + 6+ 8+......n-1 terms )
So tn = 3 + (n-1)/2 [ 2 + 2 + (n-1-1)*2 ]
=3 + (n-1)*2n/2 = 3 + n(n-1)
So the nth term of the series is
Tn = [ 3 + n(n-1) ] / n! = 3/n! + n(n-1)/n! = 3/n! + 1/(n-2)!
So sum = 3  1/n! + 1/(n-2)! = 3(e-1) + e = 4e -3
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