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Find the sum of the following series. Round to the nearest hundredth if necessary,
3 + 6 + 12+ ... +49152
Sum of a finite geometric series:
Q] - a
Sn =
1-T

Sagot :

Answer:

[tex]sum = \frac{a( {r}^{n - 1} )}{r - 1} \\ : but \: l = a( {r}^{n - 1} ) \\ 49152 = 3( {2}^{n - 1} ) \\ 16384 = {2}^{n - 1} \\ {2}^{n} = 32768 \\ {2}^{n} = {2}^{15} \\ n = 15 \\ \therefore \: sum = \frac{3(2 {}^{15 - 1}) }{15 - 1} \\ = \frac{49152}{14} \\ = 3510.9[/tex]

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