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

1 Answer
Feb 11, 2018

Answer:

# \ #

# \qquad \qquad \qquad \qquad \qquad \qquad sum_{i=0}^{10} 8 ( - 1/4 )^{ i - 1 } \ = \ - { 209715 } / { 8192 } . #

Explanation:

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# "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 !" #