已知数列an的通项公式an=(2n-1)+1/2的n次方,求Sn
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![已知数列an的通项公式an=(2n-1)+1/2的n次方,求Sn](/uploads/image/z/677163-3-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%3D%EF%BC%882n-1%EF%BC%89%2B1%2F2%E7%9A%84n%E6%AC%A1%E6%96%B9%2C%E6%B1%82Sn)
已知数列an的通项公式an=(2n-1)+1/2的n次方,求Sn
已知数列an的通项公式an=(2n-1)+1/2的n次方,求Sn
已知数列an的通项公式an=(2n-1)+1/2的n次方,求Sn
分组求和
Sn=a1+a2+a3+……+an
=(1+1/2)+(3+1/4)+(5+1/8) +……+[(2n-1)+1/2^n]
=(1+3+5+……+(2n-1))+( 1/2+1/4+1/8+……+1/2^n)
=n(1+2n-1)/2+ 1/2(1-1/2^n)/(1-1/2)
=n^2+1-1/2^n.
采用Sn-q倍Sn,错位相减法!
an=(2n-1)*(1/2)^n
Sn=1*(1/2)+3*(1/2)^2+5*(1/2)^3+……+(2n-1)*(1/2)^n
0.5Sn=1*(1/2)^2+3*(1/2)^3+……+(2n-3)*(1/2)^n+(2n-1)*(1/2)^(n+1)
两式相减:
0.5Sn=1*(1/2)+2*(1/2)^2+...
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采用Sn-q倍Sn,错位相减法!
an=(2n-1)*(1/2)^n
Sn=1*(1/2)+3*(1/2)^2+5*(1/2)^3+……+(2n-1)*(1/2)^n
0.5Sn=1*(1/2)^2+3*(1/2)^3+……+(2n-3)*(1/2)^n+(2n-1)*(1/2)^(n+1)
两式相减:
0.5Sn=1*(1/2)+2*(1/2)^2+2*(1/2)^3+……+2*(1/2)^n-(2n-1)*(1/2)^(n+1)
Sn=1+4*[(1/2)^2+(1/2)^3+……+(1/2)^n]-2(2n-1)*(1/2)^(n+1)
=3-4*(1/2)^n+(2n-1)*(1/2)^n
=(2n-5)*(1/2)^n+3
希望采纳,不懂,请追问,祝愉快
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