Question

设数列{an}的前n项和为sn,且a1=1,s(n+1)=4an+2

（1）设bn=an/2^n,求证bn是等差数列

（2）求数列an的通项公式和前n项和的公式

Answer 1

1.由题意可得S2=4A1+2

==>A1+A2=4A1+2 ==>A2=5

因为S[n+1]=4An+2 则有Sn=4A[n-1]+2

两式子相减可得 S[n+1]-Sn=4An-4A[n-1]

因为 S[n+1]-Sn=A[n+1]

所以A[n+1]=4An-4A[n-1]

==>A[n+1]-2An=2(An-2A[n-1])

所以{An-2A[n-1]}是以A2-2A1=3为首项的 2为公比的等比数列

所以An-2A[n-1]=3×2^(n-1)

==>An/2^n-A[n-1]/2^(n-1)=3/4 两边同时除以2^(n+1)

即Bn-B[n-1]=3/4 所以Bn是等差数列

2.因为B1=A1/2=1/2 ==>Bn=1/2+3/4×(n-1)=(3n-1)/4

所以An/2^n=(3n-1)/4 ==>An=(3n-1)×2^(n-2)

所以Sn=2×2^(-1)+5×2^0+8×2+..+(3n-4)×2^(n-3)+(3n-1)×2^(n-2)

2Sn=2×2^0+5×2+8×2^2+..+(3n-4)×2^(n-2)+(3n-1)×2^(n-1)

Sn=(3n-1)×2^(n-1)+(2-5)×2^0+(5-8)×2+...+[(3n-4)-(3n-1)]×2^(n-2)-1

=(3n-1)×2^(n-1)-3×[2^0+2+2^2+...+2^(n-2)]-1

=(3n-1)×2^(n-1)-3×[2^(n-1)-1]-1

=(3n-4)×2^(n-1)+2

Answer 2

无解