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zmey [24]
3 years ago
11

Find the sum of the following series. Round to the nearest hundredth if necessary,

Mathematics
1 answer:
Aneli [31]3 years ago
3 0

Answer:

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

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