x=an y=a(n+1)代入2x-y+1=0
2an -a(n+1)+1=0
a(n+1)=2an +1
a(n+1) +1=2(an +1)
[a(n+1)+1]/(an +1)=2,為定值.
a1 +1=1+1=2
數(shù)列{an +1}是以2為首項,2為公比的等比數(shù)列.
an +1=2×2^(n-1)=2ⁿ
an=2ⁿ-1
cn=n(3an+2)=n[3×(2ⁿ-1)+2]=3n×2ⁿ -n
數(shù)列{cn}的通項公式為cn=3n×2ⁿ -n.