来自黄效国的问题
1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+……+60+61)正确结果是1又31分之30
1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+……+60+61)
正确结果是1又31分之30
1回答
2020-03-24 20:31
1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+……+60+61)正确结果是1又31分之30
1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+……+60+61)
正确结果是1又31分之30
1+2=2*3/2
1+2+3=3*4/2
1+2+3+4=4*5/2
……………………
1+2+3+……+61=61*62/2.1+2+3+……+n=n(n+1)/2,这是一个等差数列求和的公式.所以,原式=1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+61)
=1+2/(2*3)+2/(3*4)+2/(4*5)+……+2/(61*62)=2[(1/2+1/(2*3)+1/(3*4)+1/(4*5)+……+1/(61*62)〕=2[(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+……+(1/61-1/62)]=2(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+……+1/61-1/62)=2(1-1/62)=61/31=1又31分之30.