EvilZone
Encyclopedia Galactica => Science => : z3ro February 09, 2013, 04:19:32 PM
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[Question]
What's the sum of 1 + 2 + 4 + 8 + 16 + ... (continues to infinity with each number twice as large as the previous one)
Sum? ??? = Infinity ∞ ?! Right?
....
But wait.... :o
1 * X = X for any X [OK?]
so the sum is similar to (1)(1+2+4+8+16+...) [OK?]
Now, One is the same as (two minus one)
1 = (2-1) [OK?]
So.. the sum is still the same as (2-1)(1+2+4+8+16+...)
If we multiply..... ??? we get... ( 2+4+8+16+32+...... -1 + -2 + -4 + -8 + -16 + -32 +...) We have
2 - 2 = 0
+4 + -4 =0
...etc...
Every number cancels out.. leaving... -1 8)
(1+2+4+8+16+...) = -1
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no.... i'm not sure if this is a troll...
.. See, your cancellation has a flaw.
you are canceling the SECOND term of the negative series with first of the positive...
hence sum is
∞ - 1 = ∞ ...
you cannot forget the un-cancelled term.
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So.. the sum is still the same as (2-1)(1+2+4+8+16+...)
If we multiply..... ??? we get... ( 2+4+8+16+32+...... -1 + -2 + -4 + -8 + -16 + -32 +...) We have
2 - 2 = 0
+4 + -4 =0
...etc...
Every number cancels out.. leaving... -1 8)
(1+2+4+8+16+...) = -1
Your logic is flawed. If you multiply:
2[(2-1)(1+2+4+8+16+...)] =
2(2-1)*2(1+2+4+8+16+...) =
(4-2)*(2+4+8+16+32+...)
which is still infinity.
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You put too many spaces out of excitement.
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You put too many spaces out of excitement.
Lmfao. Exactly.
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You can't just write one equation, take a part of it and rewrite it to suite your needs. Well... sometimes you can but this is not the case. Rules must be followed and (2-1)*(1+2+4+8+...) is still equal to ∞ no matter what.
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You are not multiplying, subtracting, or dividing. You are also not adding by a negative, which means the answer will always be a positive. The answer is always going to be infinite.