a1=S1=2
當(dāng)n≥2時(shí),an=Sn-S(n-1)=n²+1-(n-1)²-1=2n-1
則b1=2/3
當(dāng)n≥2時(shí),bn=2/(an+1)=2/(2n-1+1)=1/n
Tn=2/3+1/2+1/3+……+1/n
Cn=T(2n+1)-Tn=1/(n+1)+1/(n+2)+……+1/(2n+1)
當(dāng)n=k時(shí),
Ck=1/(k+1)+1/(k+2)+……+1/(2k+1)
當(dāng)n=k+1時(shí),
C(k+1)=1/(k+2)+1/(k+3)+……+1/(2k+1)+1/(2k+2)+1/(2k+3)
C(k+1)-Ck=1/(k+2)+1/(k+3)+……+1/(2k+1)+1/(2k+2)+1/(2k+3)-[1/(k+1)+1/(k+2)+……+1/(2k+1)]
=1/(2k+2)+1/(2k+3)-1/(k+1)
=1/(2k+3)-1/(2k+2)