Question

已知数列an满足a1=b,an+1=can+n+1,其中b,c是常数,且c不等于1,求通项an

Answer 1

an+1 +X=C[an+(n+1+X)/C]

∴X=(n+1+X)/C

解得,X=(n+1)/c-1

[an+1 + (n+1)/c-1]/[an+(n+1)/c-1]=C

∴数列Kn=an+(n+1)/c-1 公比为C

K1=b+2/(c-1) 所以Kn=K1*q^n-1

Kn=an+(n+1)/(c-1)=[b+2/(c-1) ]*C^n-1

an=[b+2/(c-1) ]*C^n-1 -(n+1)/(c-1)