已知数列An的前n项和为Sn,若An=1+1/(根号n+根号n+2)求a1+a2+ +an
1个回答
展开全部
a(n)=1-1/[√n+√(n+2)]
=1-[√(n+2)-√n]/[(n+2)-n]
=1-(1/2)[√(n+2)-√n]
所以
S(n)
=a(1)+a(2)+…+a(n)
=1-(1/2)(√3-1)+1-(1/2)(√4-√2)+1-(1/2)(√5-√3)+1-(1/2)(√6-√4)+…+1-(1/2)[√(n+1)-√(n-1)]+1-(1/2)[√(n+2)-√n]
=n×1-(1/2)[√3-1+√4-√2+√5-√3+√6-√4+…+√(n+1)-√(n-1)+√(n+2)-√n]
=n-(1/2)[-1-√2+√(n+1)+√(n+2)]
=n+(1/2)[1+√2-√(n+1)-√(n+2)]
=1-[√(n+2)-√n]/[(n+2)-n]
=1-(1/2)[√(n+2)-√n]
所以
S(n)
=a(1)+a(2)+…+a(n)
=1-(1/2)(√3-1)+1-(1/2)(√4-√2)+1-(1/2)(√5-√3)+1-(1/2)(√6-√4)+…+1-(1/2)[√(n+1)-√(n-1)]+1-(1/2)[√(n+2)-√n]
=n×1-(1/2)[√3-1+√4-√2+√5-√3+√6-√4+…+√(n+1)-√(n-1)+√(n+2)-√n]
=n-(1/2)[-1-√2+√(n+1)+√(n+2)]
=n+(1/2)[1+√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?
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