How do you find the sum of the finite geometric sequence of #Sigma 8(-1/4)^(i-1)# from i=0 to 10?

Question

Evelyn Agard · Accepted Answer

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# \qquad \qquad \qquad \qquad \qquad \qquad sum_{i=0}^{10} 8 ( - 1/4 )^{ i - 1 } \ = \ - { 209715 } / { 8192 } . # # \ # # "Let's recall the general formula for the sum of a finite geometric" # # "series:" # # sum_{i=0}^{n} a r^n \ = \ { a ( 1 - r^n ) } / (1 - r ); \qquad \quad "first term" = 1, \ "common ratio" = r. # # "In order to apply this to our example, let's rewrite the original:" # # "Given:" # # sum_{i=0}^{10} 8 ( - 1/4 )^{ i - 1 } \ = \ 8 \ sum_{i=0}^{10} ( - 1/4 )^{ i - 1 } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 8 \ sum_{i=0}^{10} ( - 1/4 )^{- 1 }( - 1/4 )^{ i } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 8 ( - 1/4 )^{- 1 } \ sum_{i=0}^{10} ( - 1/4 )^{ i } # # "Applying the general formula above to the latter sum, we note" # # "that we should let:" \qquad n = 10, a = 1, r = (- 1/4)." # #"And now, continuing, we substitute these into that formula," # # "and then simplify afterward:" # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 8 ( - 1/4 )^{- 1 } \cdot { 1 \cdot ( 1 - (- 1/4)^10 ) } / (1 - (- 1/4) ) # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 8 ( - 4 ) \ { ( 1 - (- 1/4)^10 ) } / (1 - (- 1/4) ) # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 8 ( - 4 ) \ { ( 1 - (- 1/4)^10 ) } / (1 - (- 1/4) ) \cdot { (-4)^10 }/{ (-4)^10 } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ -32 \ { ( (-4)^10 - (1)^10 ) } / ( (-4)^10 - (-4)^9 )# # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ -32 \ { (-4)^10 - 1 } / { (-4)^10 - (-4)^9 } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ -32 \ { 4^10 - 1 } / { 4^10 - (-4^9) } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ -32 \ { 4^10 - 1 } / { 4^10 + 4^9 } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ - 2^5\ { (2^2)^10 - 1 } / { (2^2)^10 + (2^2)^9 } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ - 2^5\ { 2^20 - 1 } / { 2^20 + 2^18 } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ - \color(red)cancel{ 2^5 }\ { 2^20 - 1 } / { \color(red)cancel{ 2^5 } ( 2^15 + 2^13 ) } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ - { 2^20 - 1 } / { ( 2^15 + 2^13 ) } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ - { 2^20 - 1 } / { 2^13 ( 2^2 + 1 ) } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ - { 2^20 - 1 } / { 5 \cdot 2^13 } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ - { 1048576 - 1 } / { 5 \cdot 8192 } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ - { 1048575 } / { 5 \cdot 8192 } # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ - { 209715 } / { 8192 } . # # \ # # "Thus:" \qquad \qquad \qquad \qquad \qquad sum_{i=0}^{10} 8 ( - 1/4 )^{ i - 1 } \ = \ - { 209715 } / { 8192 } . # # \ # # "My sincerest apologies for all the machinations in the" # # "simplification. It's not really that bad, if you look at it." # # "I had no idea it would be so long !! Sorry again !" #

Christopher Allers · Answer

undefined To find the sum of the finite geometric sequence $ \sum_{i=0}^{10} 8 \left(-\frac{1}{4}\right)^{i-1} $:

1. Determine the first term ($ a $) and the common ratio ($ r $) of the geometric sequence.
2. Use the formula for the sum of a finite geometric series:

$$ S_n = \frac{a(1 - r^n)}{1 - r} $$

Given $ a = 8 $ and $ r = -\frac{1}{4} $:

$$ S_{10} = \frac{8 \left(1 - \left(-\frac{1}{4}\right)^{10}\right)}{1 - \left(-\frac{1}{4}\right)} $$

$$ S_{10} = \frac{8 \left(1 - \frac{1}{4^{10}}\right)}{1 + \frac{1}{4}} $$

$$ S_{10} = \frac{8 \left(1 - \frac{1}{1048576}\right)}{\frac{5}{4}} $$

$$ S_{10} = \frac{8 \times \frac{1048575}{1048576}}{\frac{5}{4}} $$

$$ S_{10} = \frac{8388600}{5242875} $$

$$ S_{10} \approx 1.599999999923428 $$

Therefore, the sum of the finite geometric sequence $ \sum_{i=0}^{10} 8 \left(-\frac{1}{4}\right)^{i-1} $ is approximately $ 1.599999999923428 $.