数列｛an｝中,a1=2,an+1=4an-3n+1,求数列Sn,证明不等式Sn+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