標題:
數學知識交流---求值(2)
發問:
(1) 求 1 - 2 + 4 - 8 + 16 - 32 + ... 的值。
最佳解答:
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)
其他解答:
數學知識交流---求值(2)
發問:
(1) 求 1 - 2 + 4 - 8 + 16 - 32 + ... 的值。
最佳解答:
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)
其他解答:
此文章來自奇摩知識+如有不便請留言告知
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) + ...... = - ( 4? + 41 + 42 + 43 + ...... ) = - ∞
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