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OpenStudy (anonymous):
help please!
Evaluate the summation of 20 times 0.5 to the n minus 1 power, from n equals 3 to 12..
9.99
19.95
29.98
39.96
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OpenStudy (anonymous):
\[\sum_{n=3}^{12} 20 \times 0.5^{n-1}\]
This is it, right?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
@larryboxaplenty
OpenStudy (anonymous):
I can't help you right now, sorry, g2g for a sec
someone else can help. I know the answer for forget the shortcut
OpenStudy (anonymous):
can someone please help me
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OpenStudy (jhannybean):
\[\large 20 \cdot \left(0.5^{n-1}\right)\] start with plugging in \(n=3\) and work your way up to \(n=12\) by adding all the values.
OpenStudy (jhannybean):
\[\large 20\cdot (0.5^{3-1} + 0.5^{4-1}+0.5^{5-1}+...+0.5^{12-1})\]
OpenStudy (welshfella):
or you can use the formula for the sum of a geometric series
Sn = a . (1 - r^n)
------
1 - r
OpenStudy (welshfella):
here a = 20*0.5^2 , n = 11 and r = 0.5
OpenStudy (anonymous):
so would i get 9.99 as the answer
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OpenStudy (anonymous):
@welshfella
OpenStudy (welshfella):
yes
OpenStudy (anonymous):
thank you
OpenStudy (welshfella):
yw
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