数学,求和 k⼀(2^k) . (k=1,2...) 求解~~

2025-03-13 14:15:55
推荐回答(2个)
回答1:

设和为S
则:S=1/2+2/2^2+3/2^3+......+K/2^K.................................(1)
S/2= 1/2^2+2/2^3+.......+(K-1)/2^K+K/2^(K+1)..........(2)
(1)-(2):
-S/2=1/2+1/2^2+1/2^3+.......+1/2^K-K/2^(K+1)
=1-1/2^K-1/2^(K+1)
=1-3/2^(K+1)
S=3/2^K-2

回答2:

S = 1/2 + 2/2^2 + 3/2^3 + 4/2^4 + ...
= 1/2 + 1/2^2 + 1/2^2 + 1/2^3 + 2/2^3 + 1/2^4 + 3/2^4 + ....
= 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + .... + (1/2)(1/2 + 2/2^2 + 3/2^3 + ... )
= (1/2) / (1-1/2) + (1/2)S
S - (1/2)S = 1
S = 2
Answer 2