1/(1*2)+1/(2*3)+1/(3*4)+……+1/(19*20)=用简便算法写出过程快!谢谢

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 03:57:31
1/(1*2)+1/(2*3)+1/(3*4)+……+1/(19*20)=用简便算法写出过程快!谢谢

1/(1*2)+1/(2*3)+1/(3*4)+……+1/(19*20)=用简便算法写出过程快!谢谢
1/(1*2)+1/(2*3)+1/(3*4)+……+1/(19*20)=用简便算法写出过程
快!谢谢

1/(1*2)+1/(2*3)+1/(3*4)+……+1/(19*20)=用简便算法写出过程快!谢谢
1/1*2+1/2*3+1/3*4+1/4*5+1/5*6.1/19*20
=(1-1/2)+(1/2-1/3)+(1/3-1/4)…+(1/18-1/19)+(1/19-1/20)
=1-1/20
=19/20

原式=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...(1/19-1/20)
=1-1/20
=19/20

1/(1*2)+1/(2*3)+1/(3*4)+……+1/(19*20)
=1-1/2+1/2-1/3+1/3-1/4+...+1/19-1/20
=1-1/20
=19/20
=0.95

1+(1+2分之一)+(1+2+3)分之一+(1+2+3+4)分之一+...+(1+2+...+2013)分之一 (1+1/2)(1+1/3)(1+1/4)(1+1/5)……(1+1/2005)(1+1/2006)=? 1+(1+2)+(1+2+3)+…+(1+2+3+4+…+100) 及运算过程 (1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)×(1/2+1/3+1/4) (1978×1979+1979×1980)÷(1+2+3+.+1979) 请猜想1+3+5+7+9+……+(2n+1)+(2n+3)=? 1×2+2×3+……+n(n+1)= (1+2+3分之1)+(1+2+3+4分1)+~(1+2+3+4+~+2000分之1) S=0×1+1×2+2×3+3×4+…+(n-1)×n [1/(√2+1)+1/(√3+√2)+…+1/(√2015+√2014)]×(√2015+1) 2*(3+1)*(3^2+1)*(3^4+1)*……*(3^22+1)+1怎么算 1、1+2+3=2×3 2、1+2+3+4+5=3×5 3、1+2+3+4+5+6+7=( )×( ) 4、11+12+13+14+15+16+17=( )×( )=( ) 5、25+26+27+28+29+30+31+32+33=( )×( )=( )再仿照 1+(1+2)/1+(1+2+3)/1+...+(1+2+...+2013)/1等于多少?要有过程! (+1)+(-2)+(+3)+(-4)++(+99)+(-100)等于几? 用vfp设计程序,求s=1+(1+2)+(1+2+3)+…+(1+2+3+…+n)的值 求和:1+2+3+…+(n-1)的和 1/2+(1/3+2/3)+(1/4+2/4+3/4)+.+(1/40+2/40+.+38/40+39/40)1/2+(1/3+2/3)+(1/4+2/4+3/4).+(1/60+2/60+.+58/60+59/60) 3(4+1)(4^2+1)+1 平方差公式的!