数学等差数列求和难题求和:1+(1/1+2)+(1/1+2+3)+······+(1/1+2+3+·····+n)

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/28 08:01:57
数学等差数列求和难题求和:1+(1/1+2)+(1/1+2+3)+······+(1/1+2+3+·····+n)数学等差数列求和难题求和:1+(1/1+2)+(1/1+2+3)+······+(1/

数学等差数列求和难题求和:1+(1/1+2)+(1/1+2+3)+······+(1/1+2+3+·····+n)
数学等差数列求和难题
求和:1+(1/1+2)+(1/1+2+3)+······+(1/1+2+3+·····+n)

数学等差数列求和难题求和:1+(1/1+2)+(1/1+2+3)+······+(1/1+2+3+·····+n)
你求的是1+1/(1+2)+1/(1+2+3)+······+1/(1+2+3+·····+n)吧
1+2+3+·····+n=n(n+1)/2
1/(1+2+3+·····+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
所以:
1+1/(1+2)+1/(1+2+3)+······+1/(1+2+3+·····+n)
=2(1/1-1/2)+2(1/2-1/3)+2(1/3-1/4)+……+2[1/n-1/(n+1)]
=2[1/1-1/(n+1)]
=2n/(n+1)

等差数列前n项和一写,然后用裂项法。如果没算错的话是2n/(n+1)

an=1/[n(n+1)/2]=2/[n(n+1)]=2[1/n-1/(n+1)]
Sn=2[1/1-1/(n+1)=2n/(n+1)