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∑[i=1,∞,(12)(0.25)^i]
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\[12\sum_{i=1}^{\infty}\left(\frac{1}{4}\right)^i\] is a start
you are starting at \(i=1\) not \(i=0\) right?
correct. & i am to find the sum of this infinite geometric series.
correct. & i am to find the sum of this infinite geometric series.
\[\sum_{i=1}^{\infty}r^i=\frac{r}{1-r}\] so \[\sum_{i=1}^{\infty}(\frac{1}{4})i=\frac{\frac{1}{4}}{1-\frac{1}{4}}\] \[=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\] and \[12\times\frac{1}{3}=4\]
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Thank you SO much.
yw
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