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
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/26 11:51:17
xMJ@$=cV\D\BQIPPED⌺rdEӉNg\
h/ anOc||[LI~u=dET߹4j fqʌZ.,Jr, .X.^&esl,F'ue|bW=thIiضR.0
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+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.