1/2 + 2/2^2 + 3/2^3 +...+n/2^n=2 - (n+2)/2^n
help me thanks~~
f.4 a~~math
2006-10-19 5:29 am
回答 (1)
2006-10-19 5:42 am
✔ 最佳答案
n=1,LHS = 1/2
RHS = 2 - (1 + 2)/2 = 1/2
Assume n=k is true, ie 1/2 + 2/2^2 + 3/2^3 +...+k/2^k=2 - (k+2)/2^k
when n=k+1
LHS
= 1/2 + 2/2^2 + 3/2^3 +...+(k+1)/2^(k+1)
= 1/2 + 2/2^2 + 3/2^3 +...+k/2^k+(k+1)/2^(k+1)
= 2 - (k+2)/2^k+(k+1)/2^(k+1)
= 2 - [ 2(k+2) - (k+1) ] / 2^(k+1)
= 2 - (k+3) / 2^(k+1)
= 2 - [(k+1)+2] / 2^(k+1)
收錄日期: 2021-04-23 12:35:42
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