已知Sn=1+3/2+5/4+7/6+,+(2n-1)/(2^(n-1)) 如何用错位相减法求和.
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已知Sn=1+3/2+5/4+7/6+,+(2n-1)/(2^(n-1)) 如何用错位相减法求和.
![已知Sn=1+3/2+5/4+7/6+,+(2n-1)/(2^(n-1)) 如何用错位相减法求和.](/uploads/image/z/17950378-58-8.jpg?t=%E5%B7%B2%E7%9F%A5Sn%3D1%2B3%2F2%2B5%2F4%2B7%2F6%2B%2C%2B%EF%BC%882n-1%EF%BC%89%2F%282%5E%28n-1%29%29+%E5%A6%82%E4%BD%95%E7%94%A8%E9%94%99%E4%BD%8D%E7%9B%B8%E5%87%8F%E6%B3%95%E6%B1%82%E5%92%8C.)
Sn=1+3/2+5/4+7/8+,+(2n-3)/(2^(n-2)) +(2n-1)/(2^(n-1))
1/2Sn=1/2+3/4+5/8+7/16+,+(2n-3)/(2^(n-1)) +(2n-1)/(2^n)
Sn-1/2Sn=1+(3-1)/2+(5-3)/2^2+(7-5)/2^3+,+[(2n-1)-(2n-3)]/(2^(n-1))-(2n-1)/(2^n)
1/2Sn=1+2/2+2/2^2+2/2^3+,+2/(2^(n-1))-(2n-1)/(2^n)
=1+2*[1/2+1/2^2+1/2^3+,+1/(2^(n-1)]-(2n-1)/(2^n)
=1+2*[1-1/2^(n-1)]-(2n-1)/(2^n)
Sn=2+4*[1-1/2^(n-1)]-(2n-1)/[2^(n-1)]
注:按照通项计算题中第四项应为7/8
1/2Sn=1/2+3/4+5/8+7/16+,+(2n-3)/(2^(n-1)) +(2n-1)/(2^n)
Sn-1/2Sn=1+(3-1)/2+(5-3)/2^2+(7-5)/2^3+,+[(2n-1)-(2n-3)]/(2^(n-1))-(2n-1)/(2^n)
1/2Sn=1+2/2+2/2^2+2/2^3+,+2/(2^(n-1))-(2n-1)/(2^n)
=1+2*[1/2+1/2^2+1/2^3+,+1/(2^(n-1)]-(2n-1)/(2^n)
=1+2*[1-1/2^(n-1)]-(2n-1)/(2^n)
Sn=2+4*[1-1/2^(n-1)]-(2n-1)/[2^(n-1)]
注:按照通项计算题中第四项应为7/8
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