数列an中an+1=3an+2ˆn+6bn=an+2ˆn+3a1=-2证明bn为等比数列并求an第二问。求数列an/bn的前n项和sn

Question

数列an中 an+1=3an+2ˆn+6 bn=an+2ˆn+3 a1=-2
证明bn为等比数列 并求an
第二问。求数列an/bn的前n项和sn

Answer 1

a(n+1)=3an+2ˆn+6
bn=an+2ˆn+3
b(n+1)=a(n+1)+2ˆ(n+1)+3
=3an+2ˆn+6+2ˆ(n+1)+3
=3an+2ˆn+2*2ˆn+9
=3an+3*2ˆn+9
=3(an+2ˆn+3)
b(n+1)/bn
=3(an+2ˆn+3)/(an+2ˆn+3)
=3
所以是以3为公比的等比数列
b1=a1+2^1+3
=-2+2+3
=3
bn=b1*q^(n-1)
=3*3^(n-1)
=3^n
an+2^n+3=3^n
an=3^n-2^n-3
an/bn
=(3^n-2^n-3)/3^n
=3^n/3^n-2^n/3^n-3/3^n
=1-(2/3)^n-(1/3)^(n-1)
sn
=1-(2/3)^1-(1/3)^0+(2/3)^2-(1/3)^1+.+1-(2/3)^n-(1/3)^(n-1)
=1+1+1+.+1-(2/3)^1-(2/3)^2-.-(2/3)^n-(1/3)^0-(1/3)^1-.-(1/3)^(n-1)
=n-(2/3)*[1-(2/3)^n]/(1-2/3)-[1-(1/3)^n]/(1-1/3)
=n-2*[1-(2/3)^n]-3*[1-(1/3)^n]/2
=n-2+2*(2/3)^n-3/2+3*(1/3)^n/2
=2*(2/3)^n+(1/3)^(n-1)/2+n-5/2