數學知識交流---求值(2)

Let L = 1 - 2 + 4 - 8 + 16 - 32 + ... ---(1)
Then (-2)L = - 2 + 4 - 8 +16 + ... ---(2)
(1)-(2):
=> L - (-2)L = 1
=> 3L = 1
=> L = (1/3)

1 - 2 + 4 - 8 + 16 - 32 + ...
= 1 + (-2)^1 + (-2)^2 + (-2)^3 + (-2)^4 + ...
= 1 + [(-2)^1 + (-2)^2 + (-2)^3 + (-2)^4 + ... ]
= 1 + lim n->∞ [1 - (-2)^n] / [1 - (-2)]
= 1 + lim n->∞ [1 - (-2)^n] / 3
= 1 + 1/3 lim n->∞ [1 - (-2)^n]

The answer depends on whether the number of terms is odd or even .

When n is odd,
(-2)^n -> -∞.
[1 - (-2)^n] -> +∞

1 - 2 + 4 - 8 + 16 - 32 + ...
= +∞

When n is even,
(-2)^n -> ∞.
[1 - (-2)^n] -> -∞

1 - 2 + 4 - 8 + 16 - 32 + ...
= -∞



Therefore, when there is odd number of terms, the value will be the positive infinity.
When there is even number of terms, the value will be the negative infinity

T(1) = 1
T(2) = -2
Common Ratio = -2
As |common ratio| > 1,
Hence, the value is oscillating vigorously as it tends to infinity
Therefore, 1 - 2 + 4 - 8 + 16 - 32 + ... is unknown

To 意見者002 ,

若此數式的最後一數為正數，那答案就不同了!

(1)
解：1 - 2 + 4 - 8 + 16 - 32 + ...
= ( 1 - 2 ) + ( 4 - 8 ) + ( 16 - 32 ) + ...
= (-1) + (-4) + (-16) + ......
= - ( 4⁰ + 4¹ + 4² + 4³ + ...... )
= - ∞