ceany欣儿
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错位相减求和
Sn=1/2^1+3/2^2+5/2^3+.+(2n-3)/2^(n-1)+(2n-1)/2^n ①‘
①×1/2 (1/2)Sn= 1/2^2+3/2^3+.+(2n-3)/2^n+(2n-1)/2^(n+1) ②
①-② (1/2)Sn=1/2+ 1/2 +1/2^2+.+1/2^(n-1)-(2n-1)/2^(n+1)
=1/2+ 1-1/2^(n-1)-(2n-1)/2^(n+1)
Sn =1 + 2 - 4/2^n-(2n-1)/2^n
∴ Sn=3-(2n+3)/2^n
1年前
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