1/3^2-1+1/5^2-1+1/7^2-1...+1/(2n+1)^2-1+...=

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 01:19:26

1/3^2-1+1/5^2-1+1/7^2-1...+1/(2n+1)^2-1+...=
1/3^2-1+1/5^2-1+1/7^2-1...+1/(2n+1)^2-1+...=

1/3^2-1+1/5^2-1+1/7^2-1...+1/(2n+1)^2-1+...=
数列裂项消去法,分母分开,然后约掉就可以了

1/3+1/5+1/7+.+1/2n+1 ((1/2)-1)*((1/3)-1)*((1/4)-1)*((1/5)-1)*((1/6)-1)*((1/7)-1)*((1/8)-1)*((1/9)-1)*((1/10)-1) matlab 编程数组的数据如下:8 1 1 1 1 1 1 3 3 2 1 1 5 1 1 3 1 1 2 1 1 5 3 3 3 1 1 4 5 1 1 1 1 1 2 2 2 2 4 3 1 5 4 2 1 1 1 2 1 3 1 1 2 2 5 2 1 3 2 5 1 1 3 1 1 1 1 2 1 5 4 2 2 1 3 4 1 2 3 1 2 4 4 1 1 1 2 2 2 2 2 1 1 4 4 1 3 2 1 1 5 1 1 3 7 1 1 1+3+5+7+.+(2n-1) (1+1/2)*(1+1/4)*(1+1/6)*.*(1+1/20)*(1-1/3)*(1-1/5)+(1-1/7)*.*(1-1/2 1/3^2-1+1/5^2-1+1/7^2-1...+1/(2n+1)^2-1+...= 放缩法证明1/3^2+1/5^2+1/7^2+.+1/(2n+1)^2 求证1/(2*3)+1/(3*5)+1/(4*7)+...+1/((n+1)(2n+1)) 1/1*3=1/2(1-1/3)1/3*5=1/2(1/3-1/5)1/5*7=1/2(1/5-1/7).1/17*19=1/2(1/17-1/19)所以1/1*3+1/3*5+1/5*7+.1/17*19=1/2(1-1/3)+1/2(1/3+1/5)+1/2(1/5-1/7)+.1/2(1/17-1/19)=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7.+1/17-1/19)=1/2(1-1/1 9*(1-1/2)*(1-1/3)*(1-1/4)*(1-1/5)*(1-1/6)*(1-1/7)*(1-1/8)*(1-1/9)怎样简便计算 (1+2/1)*(1+4/1)*(1+6/1)*...*(1+20/1)*(1-3/1)*(1-5/1)*(1-7/1)*...*(1-21/1)等于多少 因为1/1*3=1/2*(1-1/3),1/3*5=1/2*(1/3-1/5),1/5*7=1/2*(1/5-1/7),.,1/17*19=1/2*(1/17-1/19)所以1/1*3+1/3*5+1/5*7+...+1/17*19=1/2*(1-1/3)+1/2*(1/3-1/5)+1/2*(1/5-1/7)+...+1/2*(1/17-1/19)=1/2*(1-1/3+1/3-1/5+1/5-1/7+...+1/17-1/19)=1/2*(1-1/19)=9/19(1 1/2+1/3+1/4+1/5+1/6+1/7+.1/20= (3+5/7-2/3)×(1/5+1/7+1/13)-(1/5-1/7+1/13)×3+(1/5+1/7+1/13)×(2/3-7/5)有一个打错了(3+5/7-2/3)×(1/5+1/7+1/13)-(1/5-1/7+1/13)×3+(1/5+1/7+1/13)×(2/3-7/5) (1/2-1/3)+(1/4-1/5)+(1/7-1/10)+(1/14-1/15)+(1/28-1/30)(1/2-1/3)+(1/4-1/5)+(1/7-1/10)+(1/14-1/15)+(1/28-1/30) 1/3 1/2 5/11 7/18 1/3 (1)1+1/2+1/3+1/4+1/5+1/6+1/7+1/14+1/28 求和:1/1*3+1/2*4+1/3*5+.+1/n(n+1) 1/1*4+1/4*7+1/7*10+.+1/(n+2)(n+1)