tmtz.net
当前位置:首页 >> 计算:(2^2+1)/(2^2%1)+(3^2+1)/(3^2%1)+…+(2005^2... >>

计算:(2^2+1)/(2^2%1)+(3^2+1)/(3^2%1)+…+(2005^2...

解答: 看通项即可 (n²+1)/(n²-1) =(n²-1+2)/(n²-1) =(n²-1)/(n²-1)+2/(n²-1) =1+2/(n-1)(n+1) =1+1/(n-1)-1/(n+1) ∴ (2^2+1)/(2^2-1)+(3^2+1)/(3^2-1)+…+(2005^2+1)/(2005^2-1). =(1+1/1-1/3)+(1+1/2-1/4)+...

(1/2+1/3)+(1/2^2+1/3^2)+……+(1/2^n+1/3^n)+...... limSn=lim[(1/2+1/3)+(1/2^2+1/3^2)+……+(1/2^n+1/3^n)] =lim[(1/2+1/2^2++……+1/2^n]+lim[1/3+1/3^2+...1/3^n] =(1/2)/(1-1/2)+(1/3)/(1-1/3) =1+1/2=3/2 故级数收敛,且和=3/2

(1^2+2^2+...+n^2)/(n^3+n)

# include # include int main (){ double sum; int sum1=0; int i; for (i=1;i

>> N = 200;>> x = linspace(1,N,N);>> y = sum(1./(2*x-1).^2)y = 1.2325

代码只有两行: n=1:100; test=sqrt(6*sum(1./(n.*n))) 测试结果: n=100时,结果= 3.1321; n=1000时,结果= 3.1406; n=10000时,结果= 3.1415: 简单解释一下:这是用向量的方法求解,效率最高,比用循环快的多 test=sum(1./(n.*n)) 包含好几步...

1^3+2^3+3^3+……+n^3=[n(n+1)/2]^2 证明: 利用立方差公式: (n+1)^4-n^4=[(n+1)^2+n^2][(n+1)^2-n^2] =(2n^2+2n+1)(2n+1) =4n^3+6n^2+4n+1 2^4-1^4=4*1^3+6*1^2+4*1+1 3^4-2^4=4*2^3+6*2^2+4*2+1 4^4-3^4=4*3^3+6*3^2+4*3+1 ...... (n+1)^4-n^4=...

1+(1+2)+(1+2+3)+....+(1+2+3+4+...+99) //前99项之和 =(100-1)*1+(100-2)*2+(100-3)*3+...+(100-99)*99 //99个1,98个2,97个3,。。1个99 =100(1+2+..+99)-(1^2+2^2+3^2+..+99^2) =100*[99*(1+99)/2]-99*(99+1)*(2*99+1)/6 最终结果=100*99*(5...

1/n*(n+1)=1/n-1/(n+1); (1/1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/2003-1/2004)+(1/2004-1/2005) =1-1/2005=2004/2005;

(1+3)(1+3^2)(1+3^4)……(1+3^64)+1 =1/2*2*(1+3)(1+3^2)(1+3^4)……(1+3^64)+1 =1/2*(3-1)*(1+3)(1+3^2)(1+3^4)……(1+3^64)+1 =1/2*(3^2-1)(1+3^2)(1+3^4)……(1+3^64)+1 =1/2*(3^4-1)(1+3^4)……(1+3^64)+1 =1/2*(3^8-1)……(1+3^64)+1 .......... =1/2*(...

网站首页 | 网站地图
All rights reserved Powered by www.tmtz.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com