問題1:2a1=S1+2=a1+2
→a1=2
2a2=S2+2=a1+a2+2
→a2=4
問題2:an是Sn與2的等差中項(xiàng)
→2an=Sn+2
2a(n-1)=S(n-1)+2
兩式相減
2an-2a(n-1)=Sn-S(n-1)=an
→an=2a(n-1)
→an=2a(n-1)=2^2a(n-2)=……＝2^(n-1)a1=2^n
點(diǎn)P(bn,bn+1)在直線y=x+2上
→b(n+1)=bn+2
→bn=b(n-1)+2=b(n-2)+2*2=b(n-3)+2*3=……=b1+2(n-1)=2+2(n-1)=2n
問題3:設(shè)Cn=an*bn=n*2^(n+1)
Tn=1*2^2+2*2^3+3*2^4+……+n*2^(n+1)
→2Tn= 1*2^3+2*2^4+……+(n-1)*2^(n+1)+n*2^(n+2)
兩式相減
-Tn=Tn-2Tn=1*2^2+1*2^3+1*2^4+……+1*2^(n+1)-n*2^(n+2)
=4+8+16+……+2^(n+1)-n*2^(n+2)
=4+4+8+16+……+2^(n+1)-n*2^(n+2)-4
=8+8+16+……+2^(n+1)-n*2^(n+2)-4
=……
=2^(n+2)-n*2^(n+2)-4
=-(n-1)*2^(n+2)-4
故Tn=(n-1)*2^(n+2)+4