1.求通项公式为an=(2n+1)/{n(n+1)(n+2)}的数列前n项和Sn 2.数列{an}满足an=a（n-3）,a1=1,a2=3,a3=5,求Sn
1.求通项公式为an=(2n+1)/{n(n+1)(n+2)}的数列前n项和Sn
2.数列{an}满足an=a（n-3）,a1=1,a2=3,a3=5,求Sn
两道题目 谢谢谢谢。
=2/(n+1) - 2/(n+2) + (1/2)/[n(n+1)] - (1/2)/[(n+1)(n+2)],
s(n)=2[1/2-1/3 + 1/3-1/4 + ... + 1/(n+1)-1/(n+2)] +(1/2)[1/[1*2] - 1/[2*3] + 1/[2*3] - 1/[3*4] + ... +1/[n(n+1)] - 1/[(n+1)(n+2)]
=2{1/2 - 1/(n+2)] + (1/2)[1/[1*2] - 1/[(n+1)(n+2)]}
=1-2/(n+2) + 1/4 - 1/[2(n+1)(n+2)]
=5/4 - [4n+5]/[2(n+1)(n+2)]
a(n)=a(n-3),
a(n+3)=a(n),
a(3n-2)=a(1)=1,
a(3n-1)=a(2)=3,
a(3n)=a(3)=5.
s(3n)=[a(1)+a(2)+a(3)]+...+[a(3n-2)+a(3n-1)+a(3n)]=[1+3+5]n=9n.
s(3n-1)=s(3n)-a(3n)=9n-5.
s(3n-2)=s(3n-1)-a(3n-1)=9n-5-3=9n-8.本回答被提问者和网友采纳
=1/[n(n+2)]+1/(n+2)*(n+1)
=1/2[1/n-1/(n+2)]+[1/(n+1)-1/(n+2)]
Sn=1/2(1-1/3)+1/2-1/3+1/2(1/2-1/4)+1/3-1/4+...+1/2[1/n-1/(n+2)]+[1/(n+1)-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]+1/2-1/(n+2)
=5/4-1/[2(n+1)]-3/[2(n+2)]
2,an=a（n-3）
a(3n-2)=a1
a(3n-1)=a2
a(3n)=a3
Sn当n=3K+1时,K∈N
Sn=K(a1+a2+a3)+a1=9K+1
Sn当n=3K+2时,K∈N
Sn=K(a1+a2+a3)+a1+a2=9K+4
Sn当n=3K+3时,K∈N
Sn=K(a1+a2+a3)+a1+a2+a3=9K+9
肯定正确,望采纳
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