已知数列{An}满足=2An-1+2^n-1(n属于正整数,n大于等于2)且A4=81.求数列{An}的前n项和Sn
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已知数列{An}满足=2An-1+2^n-1(n属于正整数,n大于等于2)且A4=81.求数列{An}的前n项和Sn
![已知数列{An}满足=2An-1+2^n-1(n属于正整数,n大于等于2)且A4=81.求数列{An}的前n项和Sn](/uploads/image/z/6464927-47-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7BAn%7D%E6%BB%A1%E8%B6%B3%3D2An-1%2B2%5En-1%28n%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%2Cn%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E2%29%E4%B8%94A4%3D81.%E6%B1%82%E6%95%B0%E5%88%97%7BAn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
用递推公式
a4=2a3+2^4-1=81 a3=33=2a2+2^3-1
so a2=13=2a1+2^2-1
so a1=5
an=2an-1+2^n-1
=2[2an-2+2^(n-1)-1]+2^n-1
=2^2an-2+(2^n+2^n)-(2+1)
=2^2(2an-3+2^(n-2)-1)++(2^n+2^n)-(2+1)
=2^3an-3+(2^n+2^n+2^n)-(2^2+2+1)
.
.以此类推
=2^(n-1)*a1+(n-1)*2^n-[(2^(n-2)+2^(n-3).+2+1]
=5*2^(n-1)+(2n-2)*2^(n-1)-(1-2^(n-1))/(1-2)
=5*2^(n-1)+(2n-2)*2^(n-1)-2^(n-1)+1
=(5+2n-2-1)*2^(n-1)+1
=(2n+2)*2^(n-1)+1
=(n+1)*2^n+1
所以an=(n+1)*2^n+1
令an=bn+1 bn=(n+1)*2^n
bn前n项和为Tn
则Sn=Tn+n
Tn=2*2+3*2^2+4*2^3.(n+1)*2^n
2*Tn=2*2^2+3*2^3.n*2^n+(n+1)*2^(n+1)
以上两式相减得-Tn=2+2^2+2^3+.2^n-(n+1)*2^(n+1)
-Tn=2*(1-2^n)/(1-2)-(n+1)*2^(n+1)
=-n*2^(n+1)-2
so Tn=n*2^(n+1)+2
Sn=Tn+n=n*2^(n+1)+n+2
so Sn=n*2^(n+1)+n+2
哇...终于码完了=.=.=累死我廖!给点辛苦分吧.
a4=2a3+2^4-1=81 a3=33=2a2+2^3-1
so a2=13=2a1+2^2-1
so a1=5
an=2an-1+2^n-1
=2[2an-2+2^(n-1)-1]+2^n-1
=2^2an-2+(2^n+2^n)-(2+1)
=2^2(2an-3+2^(n-2)-1)++(2^n+2^n)-(2+1)
=2^3an-3+(2^n+2^n+2^n)-(2^2+2+1)
.
.以此类推
=2^(n-1)*a1+(n-1)*2^n-[(2^(n-2)+2^(n-3).+2+1]
=5*2^(n-1)+(2n-2)*2^(n-1)-(1-2^(n-1))/(1-2)
=5*2^(n-1)+(2n-2)*2^(n-1)-2^(n-1)+1
=(5+2n-2-1)*2^(n-1)+1
=(2n+2)*2^(n-1)+1
=(n+1)*2^n+1
所以an=(n+1)*2^n+1
令an=bn+1 bn=(n+1)*2^n
bn前n项和为Tn
则Sn=Tn+n
Tn=2*2+3*2^2+4*2^3.(n+1)*2^n
2*Tn=2*2^2+3*2^3.n*2^n+(n+1)*2^(n+1)
以上两式相减得-Tn=2+2^2+2^3+.2^n-(n+1)*2^(n+1)
-Tn=2*(1-2^n)/(1-2)-(n+1)*2^(n+1)
=-n*2^(n+1)-2
so Tn=n*2^(n+1)+2
Sn=Tn+n=n*2^(n+1)+n+2
so Sn=n*2^(n+1)+n+2
哇...终于码完了=.=.=累死我廖!给点辛苦分吧.
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