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

年金现值。已知年金求现值。 具体可以参考资产评估考试,资产评估考试科目第二章内容。 给你上传的文件是哪章的讲义。 P...

(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

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)+...

# include # include int main (){double sum;int sum1 = 0;double sum2 = 0;double sum3 = 0;int i;int j = 1;int t = 1;double j2;for (i=1;i

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...

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

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

方法不止一种,下面提供 2 个方法: . (1) 积分比较法: 设 P = ln[ 1/(x^2) + 1 ] 从 1 到 ∞ 的积分, 运用分部积分法,可得 P = π/2 - ln(2) = 0.8776 < 1 所以 ln(1/2^2+1)+ln(1/3^2+1)+...+ln(1/n^2+1)+... ≤ P < 1 . (2) 函数对比法: 设 f...

直接设x=tanu,u=arctan x, tan²u+1=(sin²u+cos²u)/cos²u =1/cos²u=sec²u, ⅾtanu=tan’udu =(1/cos²u)du=sec²uⅾu, ∴∫{1/[(x²+1)½]³}dx =∫1/sec³u×sec²udu =∫co...

Private Sub Command1_Click() Dim i%, s! i = 1 Do While 1 / i ^ 2 > 10 ^ -5 s = s + 1 / i ^ 2 i = i + 1 Loop Print s End Sub

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