这种题目是典型的裂项求和Cn=2/(3n^2+3n) =(2/3)*1/(n²+n) =(2/3)*1/[n(n+1)] =(2/3)*[1/n-1/(n+1)]∴ {Cn}的前n项和Tn=C1+C2+C3+.+Cn∴ Tn=(2/3)*[1-1/2+1/2-1/3+1/3-1/4+1/n-1/(n+1)]=(2/3)*[1...