求数列 {aN}:1, 1/(1+2), 1/(1+2+3),~~1/(1+2+3~~+N)。的前N项和.
来源:百度知道 编辑:UC知道 时间:2024/06/07 11:58:49
an=2/n(1+n)=2(1/n-1/(n+1)) n>=2
sn=a1+a2+……+an=1+2(1/2-1/3+1/3-1/4+……+1/n-1/(n+1))=
1+2(1/2-1/(n+1))=1+(n+1)/(n-1)=2n/(n+1)
裂项法求和:
∵1+2+...+n=n(n+1)/2
∴和式的通项公取为an=2/n(n+1)=2[(1/n)-1/(n+1)]
∴1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+n)
=2(1-1/2)+2(1/2-1/3)+2(1/3-1/4)+...+2[(1/n)-1/(n+1)]
=2[1-1/2+1/2-1/3+1/3-1/4+...+(1/n)-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
1+1/(1+2)+1/(1+2+3)+......+1/(1+2+3...+n)
=1+2/(2(2+1))+2/(3(3+1))+......2/(n(n+1))
=1+2{1/2-1/3+1/3-1/4+......+1/n-1/(n+1)}
=1+2(1/2-1/(n+1))
=2-2/(n+1)
用到的公式:1+2+3+...+N=N(N+1)/2
1/[n(n+1)]=1/n—1/(n+1)
分母是等差数列求和公式,通项表示为:2/[n(n+1)]=2/n—2/(n+1)相邻项就会消掉,会了吧。答案应该是2-2/(n+1)吧?
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