Question

设数列an的前n项和为sn已知a1=1,s(n+1)=4an+2设bn=a(n+1)-2an

Answer 1

S(n+1)=4(An)+2Sn=4A(n-1)+2两式相减A(n+1)=S(n+1)-Sn=4An-4A(n-1)A(n+1)-4An+4A(n-1)=0A(n+1)-2An=2An-4A(n-1)=2(An-2A(n-1))S2=4A1+2=4+2=6A2=S2-A1=6-1=5A2-2A1=5-2=3{A(n+1)-2An}，即｛bn}是以3为首项，2为公比的等比数列A(n+1)-2An=3×2^(n-1)两边同除2^(n+1)A(n+1)/2^(n+1)-2An/2^(n+1)=3×2^(n-1)/2^(n+1)A(n+1)/2^(n+1)-An/2^n=3/4依此类推An/2^n-A(n-1)/2^(n-1)=3/4A(n-1)/2^(n-1)-A(n-2)/2^(n-2)=3/4……A2/2-A1/1=3/4上式相加，相同项消去An/2^n-A1/2^1=3(n-1)/4An/2^n=3(n-1)/4+1/2=(3n-1)/4An=(3n-1)×2^(n-2)

Answer 2

S(n+1)=4(An)+2Sn=4A(n-1)+2两式相减A(n+1)=S(n+1)-Sn=4An-4A(n-1)A(n+1)-4An+4A(n-1)=0A(n+1)-2An=2An-4A(n-1)=2(An-2A(n-1))S2=4A1+2=4+2=6A2=S2-A1=6-1=5A2-2A1=5-2=3{A(n+1)-2An}，即｛bn}是以3为首项，2为公比的等比数列A(n+1)-2An=3×2^(n-1)两边同除2^(n+1)A(n+1)/2^(n+1)-2An/2^(n+1)=3×2^(n-1)/2^(n+1)A(n+1)/2^(n+1)-An/2^n=3/4依此类推An/2^n-A(n-1)/2^(n-1)=3/4A(n-1)/2^(n-1)-A(n-2)/2^(n-2)=3/4……A2/2-A1/1=3/4上式相加，相同项消去An/2^n-A1/2^1=3(n-1)/4An/2^n=3(n-1)/4+1/2=(3n-1)/4An=(3n-1)×2^(n-2)