作业帮1/3^2-1+1/5^2-1+1/7^2-1...+1/(2n+1)^2-1+...=
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x)373567 ZOO06lmIӠ_`gC Ϧnx1⦗ w>oSuv=[ $iy鄾绖=|aOv/}F 1d3 KMx
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)