已知数列an满足an=n(n+2),求数列1/an的前n项和sn
展开全部
an=n(n+2),
1/an=1/2[1/n-1/(n+2)]
sn=1/a1+1/a2+...+1/an
=1/2[1-1/3+1/2-1/4+1/3-1/5+....1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
1/an=1/2[1/n-1/(n+2)]
sn=1/a1+1/a2+...+1/an
=1/2[1-1/3+1/2-1/4+1/3-1/5+....1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
更多追问追答
追问
我算的是1/2【(1+1/2+...+1/n)+(-1/3-1/4...-1/n+2】然后用前n项和算,得出1/2【n(1+1/n)/2+n(-1/3-1/n+2)/2】算出来为什么跟你们的答案不一样
追答
你算的思路是对的,楼上的答案忽略了偶数项,而我的是把偶数项放在一起了。
你的1/3没消去是不对的。你再看看我分解的。
展开全部
an=n(n+2),
an=1/[n(n+2)]=(1/2)[1/n-1/(n+2)]
所以
1/an=(1/2)[1/n-1/(n+2)]
1/a(n-1)=(1/2)[1/(n-1)-1/(n+1)]
……………………
……………………
1/a2=(1/2)[1/2-1/4]
1/a1=(1/2)[1-1/3]
相加得
Sn=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=(1/2)[3/2-(2n+3)/(n+2)(n+1)]
=3/4-(2n+3)/[(n+2)(2n+2)]
an=1/[n(n+2)]=(1/2)[1/n-1/(n+2)]
所以
1/an=(1/2)[1/n-1/(n+2)]
1/a(n-1)=(1/2)[1/(n-1)-1/(n+1)]
……………………
……………………
1/a2=(1/2)[1/2-1/4]
1/a1=(1/2)[1-1/3]
相加得
Sn=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=(1/2)[3/2-(2n+3)/(n+2)(n+1)]
=3/4-(2n+3)/[(n+2)(2n+2)]
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
1/a1+1/a2+....+1/an
=1/1×3+1/3×5+....+1/n(n+2)
=1/2×(1-1/3+1/3-1/5+1/5-1/7+.....+1/n-1/(n+2))
=1/2×(1-1/(n+2))
=(n+1)/2(n+2)
=1/1×3+1/3×5+....+1/n(n+2)
=1/2×(1-1/3+1/3-1/5+1/5-1/7+.....+1/n-1/(n+2))
=1/2×(1-1/(n+2))
=(n+1)/2(n+2)
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
an=(n+2)n
1/an=1/n(n+2)=(1/n-1/n+2)/2
sn=1/a1+1/a2+……+1/an
=(1-1/3)/2+(1/2-1/4)+……+(1/n-1/n+2)
=1/2{(1-1/3)+(1/2-1/4)+……+(1/n-1/n+2)}
=1/2 *(1+1/2-1/n+1-1/n+2)
=1/2{3/2-(2n+3)/(n+1)(n+2)}
=3/4-((2n+3)/2(n+1)(n+2)
1/an=1/n(n+2)=(1/n-1/n+2)/2
sn=1/a1+1/a2+……+1/an
=(1-1/3)/2+(1/2-1/4)+……+(1/n-1/n+2)
=1/2{(1-1/3)+(1/2-1/4)+……+(1/n-1/n+2)}
=1/2 *(1+1/2-1/n+1-1/n+2)
=1/2{3/2-(2n+3)/(n+1)(n+2)}
=3/4-((2n+3)/2(n+1)(n+2)
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
1/an=1/n(n+2)=[1/n-1/(n+2)]/2,记bn=1/an,Sn=b1+b2+……+bn=[1-1/3+1/2-1/4+1/3-1/5+...+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]/2=[1+1/2-1/(n+1)-1/(n+2)]/2=(3n^2+4n+3)/(n+1)(n+2)
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询