设数列{Cn}满足Cn=2/(3n^2+3n),求{Cn}的前n项和Tn

/>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)*[1-1/(n+1)]
=(2/3)*n/(n+1)
=2n/(3n+3)