Question

数列（错位相减法）

Answer 1

an=(2n-3)*(1/2)^(n-3)
Sn=-(1/2)^(-2)+(1/2)^(-1)+3*(1/2)^0+5*(1/2)+7*(1/2)^2+....+(2n-3)*(1/2)^(n-3)
(1/2)Sn=-(1/2)^(-1)+(1/2)^0+3*(1/2)+...+(2n-5)*(1/2)^(n-3)+(2n-3)*(1/2)^(n-2)
两式相减得
(1/2)Sn=-(1/2)^(-2)+2[(1/2)^(-1)+(1/2)^0+(1/2)+..+(1/2)^(n-3)]-(2n-3)*(1/2)^(n-2)
(1/2)Sn=-(1/2)^(-2)+2[(1-1/2)^(n-1)]/(1-1/2)-[(2n-3)*(1/2)^(n-2)]
(1/2)Sn=-(1/2)^(-2)+4[(1-1/2)^(n-1)]-[(2n-3)*(1/2)^(n-2)]
(1/2)Sn=-[(1/2)^(n-3)]-[(2n-3)*(1/2)^(n-2)]
Sn=(1-2n)*(1/2)^(n-3)

Answer 2

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