求数列Cn=(2n+1)/(2^n+1)的前n项和(请写个详细的过程,
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![求数列Cn=(2n+1)/(2^n+1)的前n项和(请写个详细的过程,](/uploads/image/z/1758712-40-2.jpg?t=%E6%B1%82%E6%95%B0%E5%88%97Cn%3D%282n%2B1%29%2F%282%5En%2B1%29%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%EF%BC%88%E8%AF%B7%E5%86%99%E4%B8%AA%E8%AF%A6%E7%BB%86%E7%9A%84%E8%BF%87%E7%A8%8B%2C)
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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