在数列｛an｝中,a1=2,an+1=4an-3n+1,n∈N+.Ⅰ证明｛an-n｝是等差数列Ⅱ求数列｛an｝的前n项和SnⅢ证明不等式(前n+1项和)Sn+1≤4Sn对任意n∈N+皆成立

在数列｛an｝中,a1=2,an+1=4an-3n+1,n∈N+.Ⅰ证明｛an-n｝是等差数列Ⅱ求数列｛an｝的前n项和SnⅢ证明不等式(前n+1项和)Sn+1≤4Sn对任意n∈N+皆成立
在数列｛an｝中,a1=2,an+1=4an-3n+1,n∈N+.
Ⅰ证明｛an-n｝是等差数列Ⅱ求数列｛an｝的前n项和SnⅢ证明不等式(前n+1项和)Sn+1≤4Sn对任意n∈N+皆成立

在数列｛an｝中,a1=2,an+1=4an-3n+1,n∈N+.Ⅰ证明｛an-n｝是等差数列Ⅱ求数列｛an｝的前n项和SnⅢ证明不等式(前n+1项和)Sn+1≤4Sn对任意n∈N+皆成立
1.
a(n+1)=4an-3n+1
a(n+1)-(n+1)=4an-4n
[a(n+1)-(n+1)]/(an-n)=4,为定值.
a1-1=2-1=1
数列{an-n}是以1为首项,4为公比的等比数列.(不是等差,是等比)
2.
an-n=1×4^(n-1)=4^(n-1)
an=n+4^(n-1)
Sn=a1+a2+...+n
=(1+2+...+n)+[1+4+...+4^(n-1)]
=n(n+1)/2 +1×(4ⁿ-1)/(4-1)
=n(n+1)/2 +4ⁿ/3 -1/4
3.
S(n+1)-4Sn
=(n+1)(n+2)/2 +4^(n+1)/3 -1/4 -2n(n+1) -4^(n+1)/3+1
=(n+1)(2-3n)/2 +3/4
n=1时,S2-4S1=2×(-1)/2 +3/4=-1/4

1、应是证明an-n为等比数列！a(n+1)=4an-3n+1a(n+1)-(n+1)=4(an-n)[a(n+1)-(n+1)]/(an-n)=4故{an-n}为等比数列2、[a(n+1)-(n+1)]/(an-n)=4及a1=2知a1-1=1故an-n=4^(n-1)an=4^(n-1)+nSn=1+4+4^2+...4^(n-1)+1+2+..+n=(4^n-1)/(4-1)+n(n+1)/...

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1、应是证明an-n为等比数列！a(n+1)=4an-3n+1a(n+1)-(n+1)=4(an-n)[a(n+1)-(n+1)]/(an-n)=4故{an-n}为等比数列2、[a(n+1)-(n+1)]/(an-n)=4及a1=2知a1-1=1故an-n=4^(n-1)an=4^(n-1)+nSn=1+4+4^2+...4^(n-1)+1+2+..+n=(4^n-1)/(4-1)+n(n+1)/2=(4^n-1)/3+n(n+1)/23、S(n+1)=[4^(n+1)-1]/3+(n+1)(n+2)/24Sn=[4^(n+1)-4]/3+4n(n+1)/24Sn-S(n+1)=[4^(n+1)-4]/3+4n(n+1)/2-{[4^(n+1)-1]/3+(n+1)(n+2)/2}=-1+2n^2+2n-n^2/2-3n/2-1=3n^2/2+n/2-2=(3n^2+n-4)/2=(3n+4)(n-1)/2因n>=1故4Sn-S(n+1)>=0即S(n+1)<=4Sn

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1)a(n+1)=4an-3n+1∴a(n+1)-(n+1)=4an-3n+1-(n+1)=4an-4n=4(an-n)∴[a(n+1)-(n+1)]/(an-n)=4, a1-1=1∴{an-n}是首项为1,公比为4的等比数列2)an-n=4^(n-1)∴an=n+4^(n-1)∴Sn=1+4^0+2+4^1+...+n+4^(n-1)=(1+2+...+n)+[4^0+4^1+...+4^(n...

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1)a(n+1)=4an-3n+1∴a(n+1)-(n+1)=4an-3n+1-(n+1)=4an-4n=4(an-n)∴[a(n+1)-(n+1)]/(an-n)=4, a1-1=1∴{an-n}是首项为1,公比为4的等比数列2)an-n=4^(n-1)∴an=n+4^(n-1)∴Sn=1+4^0+2+4^1+...+n+4^(n-1)=(1+2+...+n)+[4^0+4^1+...+4^(n-1)]=n(n+1)/2+(4^n-1)/(4-1)=n(n+1)/2+(4^n-1)/33)4Sn-S(n+1)=4[n(n+1)/2+(4^n-1)/3]-[(n+1)(n+2)/2+(4^(n+1)-1)/3]=2n(n+1)-(n+1)(n+2)/2+[4^(n+1)-4-4^(n+1)+1]/3=(n+1)[2n-(n+2)/2]-1=(n+1)(3n/2-1)-1=3n??/2+n/2-2=(3n??+n-4)/2=(n-1)(3n+4)/2∵n>=1∴n-1>=0,3n+4>0∴(n-1)(3n+4)/2>=0,即4Sn-S(n+1)>=0∴4Sn>=S(n+1)

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