Posted: 3/4/2008 4:07:08 PM EDT
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How do I add: 1+(4/5)+(16/25)+ . . . +(4/5)^11 Using that E looking symbol...? |
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Well, that's a basic geometric progression of the form a+ar+ar^2+ar^3...+ar^n. That "E looking thing" is sigma, a greek letter that means "sum". The formula for the sum of a geometric progression is Sum[n] = [a*(1-r^n)]/[1-r] where "Sum[n]" means "The sum of the terms from the first to the nth". So, in your case, a=1 and r=4/5. Sum[12] = [ 1 * ( 1 - ( 4/5 )^12 ) ] / [ 1 - (4/5) ] = 4.65640261632 ETA: I forgot to mention, there's twelve terms because the first term is really 1*(4/5)^0, the second is 1*(4/5)^1, and so on up to 1*(4/5)^11, which is the twelfth term. |
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put 1 + (the sigma (E thing)). Make sure it is in the parethesis and the one is outside Put 4/5 (this is K) on the bottom of the Sigma. Put (4/5)^11 on top of the Sigma. To the right of the sigma (by the pointy part) put (K)^(2+1) I think this is right, but I'm not quite sure. I haven't done sequences in a while, and we didn't go over problems like this a whole hell of a lot. If I remember right, this should mean that you start at 4/5, and then add 4/5^2, and then add 1 to the exponent in every step, ending at 4/5^11 Edit: if the post above me is right (it probably is) my professor was a retard. Not that I didn't have my suspicions. |