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@welshfella, this is the next problem
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evaluate \[\sum_{n=3}^{12}20(0.5)^(n-1)\]
I know the the 20(0.5)^(n-1) would be100 but i'm not sure about the 12 and sigma sign again....
oh wait actually I think I got the answer 9.997 right???
100? when n = 3 , 20(0.5)^(3-1) = 20 * 0.25 = 5
wait were are you getting the .25 from????
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ohh well I got the 5 haha I don't know what I did wrong there probably plugged it in wrong but i'm still confused on were you got to .25 from..
0.5^2
0.5^(3-1) = 0.5^2
ohhh ok
i got 9.99 as the sum
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i'm just adding it all up right now, give me like 10 seconds
did you round up? because i'm still getting 9.997
nevermind lol, can you help me with one more I promise it's the last one.
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