求数列Cn=(2n+1)/(2^n+1)的前n项和（请写个详细的过程,

求数列Cn=(2n+1)/(2^n+1)的前n项和（请写个详细的过程,
求数列Cn=(2n+1)/(2^n+1)的前n项和（请写个详细的过程,

求数列Cn=(2n+1)/(2^n+1)的前n项和（请写个详细的过程,
如图,先移项,具体如下：

错位相减法
设前n项和是Sn
则 Sn=3/2²+5/2³+ 7/2^4+.........+(2n-1)/2^n+ (2n+1)/2^(n+1) ①
①乘以1/2
(1/2)Sn= 3/2³+5/2^4+.............................+(2n-1)...

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错位相减法
设前n项和是Sn
则 Sn=3/2²+5/2³+ 7/2^4+.........+(2n-1)/2^n+ (2n+1)/2^(n+1) ①
①乘以1/2
(1/2)Sn= 3/2³+5/2^4+.............................+(2n-1)/2^(n+1)+(2n+1)/2^(n+2)
则 ①-②
(1/2)Sn=3/4+2【1/2³+1/2^4+............................+1/2^(n+1)】-(2n+1)/2^(n+2)
=3/4+【1/2²+1/2³+1/2^4+............................+1/2^n]-(2n+1)/2^(n+2)
=3/4+1/2-1/2^n-(2n+1)/2^(n+2)
∴ Sn=5/2-2/2^n-(2n+1)/2^(n+1)

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