数列｛an｝中，a1=2，an+1=4an-3n+1，求数列Sn，证明不等式Sn+1<=4Sn,对任意n属于正整数成立

Answer 1

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

所以S(n+1) -4S(n)=(n+1)(n+2)/2+(4^(n+1)-1)/3-4[n(n+1)/2 + (4^n - 1)/3]
=-(n+1)(3n-2)/2+1≤0对任意n属于正整数成立

所以不等式Sn+1≤4Sn,对任意n属于正整数成立

Answer 2

a[n+1]-(n+1)=4a[n]-4n
于是b[n]=a[n]-n为q=4的等比数列
b[1]=1,b[n]=4^(n-1)
a[n]=4^[n-1]+n
S[n]=(4^n-1)/3+n(n+1)/2
S[n+1]-4S[n]=1+(n+1)(2-3n)/2
之后就好证明了，自己算吧