SOLUTION: Please help me solve this 1-((2-1)/4)-((2^2-1)/4^2)-((2^3-1)/4^3)-((2^4-1)/4^4)- infinite

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Question 1185725: Please help me solve this
1-((2-1)/4)-((2^2-1)/4^2)-((2^3-1)/4^3)-((2^4-1)/4^4)- infinite
Found 2 solutions by greenestamps, robertb:
Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!

1 - ((2-1)/4) - ((2^2-1)/4^2) - ((2^3-1)/4^3) - ((2^4-1)/4^4) - ...

1 - 1/4 - (4-1)/16 - (8-1)/64 - (16-1)/256 - ...

3/4 - (4/16-1/16) - (8/64-1/64) - (16/256-1/256) - ...

3/4 - (4/16+8/64+16/256...) + (1/16+1/64+1/256...)

The two expressions in parentheses are infinite geometric sequences with common ratio less than 1.

4/16+8/64+16/256+... = 1/4+1/8+1/16+... = (1/4)/(1-1/2) = (1/4)/(1/2) = (1/4)(2/1) = 1/2

1/16+1/64+1/256+... = (1/16)/(1-1/4) = (1/16)/(3/4) = (1/16)(4/3) = 1/12

The sum of the sequence is then

3/4 - 1/2 + 1/12 = 9/12 - 6/12 + 1/12 = 4/12 = 1/3

ANSWER: 1/3

Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website!

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