在数列{an}中,a1=2,an=2an-1+2^n+1 求an和Sn

在数列{an}中,a1=2,an=2an-1+2^n+1 求an和Sn
在数列{an}中,a1=2,an=2an-1+2^n+1 求an和Sn

在数列{an}中,a1=2,an=2an-1+2^n+1 求an和Sn

n≥2时,
an=2a(n-1) +2^(n+1)
an -2a(n-1)=2^(n+1)
等式两边同除以2^(n+1)
an/2^(n+1) -a(n-1)/2ⁿ =1,为定值.
a1/2²=2/4=1/2
数列{an/2^(n+1)}是以1/2为首项,1为公差的等差数列.
an/2^(n+1) =1/2 +1×(n-1)=(2n -1)/2
an=2^(n+1)×[(2n-1)/2]=(2n-1)×2ⁿ
n=1时,a1=(2-1)×2=2,同样满足.
数列{an}的通项公式为an=(2n-1)×2ⁿ

an=(2n-1)×2ⁿ=n×2^(n+1) -2ⁿ
Sn=a1+a2+...+an
=[1×2²+2×2³+...+n×2^(n+1)]-(2+2²+2³+...+2ⁿ)
令Cn=1×2²+2×2³+...+n×2^(n+1)
则2Cn=1×2³+2×2⁴+...+(n-1)×2^(n+1)+n×2^(n+2)
Cn-2Cn=-Cn=2²+2³+...+2^(n+1) -n×2^(n+2)
=4×(2ⁿ-1)/(2-1) -n×2^(n+2)
=(1-n)×2^(n+2) -4
Cn=(n-1)×2^(n+2) +4
Sn=Cn -(2+2²+...+2ⁿ)
=(n-1)×2^(n+2) +4 -2×(2ⁿ-1)/(2-1)
=(2n-1)×2^(n+1) +6

两边除以2^(n+1)
an/2^(n+1)=a(n-1)/2^n+1
所以an/2^(n+1)是等差数列，d=1
所以an/2^(n+1)=a1/2^(1+1)+1*(n-1)=n-1/2
所以an=(n-1/2)*2^(n+1)
即an=(2n-1)*2^n

Sn=1*2^1+3*2^2+5*2^3+……+(2n-1)*2^n

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两边除以2^(n+1)
an/2^(n+1)=a(n-1)/2^n+1
所以an/2^(n+1)是等差数列，d=1
所以an/2^(n+1)=a1/2^(1+1)+1*(n-1)=n-1/2
所以an=(n-1/2)*2^(n+1)
即an=(2n-1)*2^n

Sn=1*2^1+3*2^2+5*2^3+……+(2n-1)*2^n
2Sn=1*2^2+3*2^3+5*2^4+……+(2n-3)*2^n+(2n-1)*2^(n+1)
Sn-2Sn
=1*2^1+(3-1)*2^2+(5-3)*2^3+……+[(2n-1)-(2n-3)]*2^n-(2n-1)*2^(n+1)
=2+2*(2^2+2^3+……+2^n)-(2n-1)*2^(n+1)
=2+2*4*[1-2^(n-1)]/(1-2)-(2n-1)*2^(n+1)
=2+8*2^(n-1)-8-(2n-1)*2^(n+1)
=-(2n-3)*2^(n+1)-6
所以Sn=(2n-3)*2^(n+1)+6

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an=2an-1+2^n+1
两边同时除以2^(n)
得到 an/2^n=a(n-1)/2^(n-1)+2
∴ an/2^n-a(n-1)/2^(n-1)=2
∴ {an/2^n}是等差数列，首项为a1/2=1,公差为2
∴ an/2^n=1+2(n-1)=2n-1
∴ an=(2n-1)*2^n
利用错位想减求Sn
S...

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an=2an-1+2^n+1
两边同时除以2^(n)
得到 an/2^n=a(n-1)/2^(n-1)+2
∴ an/2^n-a(n-1)/2^(n-1)=2
∴ {an/2^n}是等差数列，首项为a1/2=1,公差为2
∴ an/2^n=1+2(n-1)=2n-1
∴ an=(2n-1)*2^n
利用错位想减求Sn
Sn =1*2+3*2^2+5*2^3+........+(2n-3)*2^(n-1)+(2n-1)*2^n ----------------------(1)
同乘以2
2Sn = 1*2^2+3*2^3+..............................+(2n-3)*2^n+(2n-1)*2^(n+1) ------ (2)
(1)-(2)
-Sn =2+2[2^2+2^3+.......................................+2^n]-(2n-1)*2^(n+1)
-Sn=2+2*[4-2^(n+1)]/(1-2)-(2n-1)*2^(n+1)
-Sn=2+2*2^(n+1)-8-(2n-1)*2^(n+1)
Sn=6+(2n-3)*2^(n+1)

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an/2^(n+1)=a(n-1)/2^n+1
故：an/2^(n+1)是公差为1，首项=1/2的等差数列
an/2^(n+1)=1/2+(n-1)
an=2^(n+1)[1/2+(n-1)]

an=2a(n-1)+2^(n+1),从2加到n得：
Sn-S1=2Sn-2an+4(1-2^n)/(1-2)
Sn=2an+4(1-2^n)-2