已知数列{an}的前n项和为Sn满足Sn-S(n-2)=3(-1/2)^(n-1)(n大于等于3)且S1=1,则S21=
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![已知数列{an}的前n项和为Sn满足Sn-S(n-2)=3(-1/2)^(n-1)(n大于等于3)且S1=1,则S21=](/uploads/image/z/1159162-34-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%E6%BB%A1%E8%B6%B3Sn-S%EF%BC%88n-2%EF%BC%89%3D3%EF%BC%88-1%2F2%29%5E%EF%BC%88n-1%EF%BC%89%EF%BC%88n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E3%EF%BC%89%E4%B8%94S1%3D1%2C%E5%88%99S21%3D)
已知数列{an}的前n项和为Sn满足Sn-S(n-2)=3(-1/2)^(n-1)(n大于等于3)且S1=1,则S21=
已知数列{an}的前n项和为Sn满足Sn-S(n-2)=3(-1/2)^(n-1)(n大于等于3)且S1=1,则S21=
已知数列{an}的前n项和为Sn满足Sn-S(n-2)=3(-1/2)^(n-1)(n大于等于3)且S1=1,则S21=
Sn-S(n-2)=3(-1/2)^(n-1)
an+an-1=3(-1/2)^(n-1)
a1=s1=1
a3+a2=3(-1/2)^2=3*1/4
a5+a4=3(-1/2)^4=3*(1/4)^2
:
a21+a20=3(-1/2)^20=3*(1/4)^10
等式两边相加得
s21=1+3[(1/4-(1/4)^11)/(1-1/4)]=2-(1/4)^10)