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1/1*2+1/2*3+1/3*4······+1/2013*2014

你好! 完整的代码,满意请采纳,有问题再问 #include #include int main(){ int i=1; double a=2.0,b=1.0,c=0.0,sum=1;while(i

(1/2+2/3+3/4+……+2013/2014)·(1+1/2+2/3+……+2012/2013)-( 1+ 1/2 +2/3+……+2013/2014)·(1/2+2/3+……+2012/2013) 设1/2+2/3+……+2012/2013=a ∴原式=(a+2013/2014)(1+a)-(1+a+2013/2014)a =a+a²+2013/2014+2013/2014a-a-a²-2013/2014a =2...

解:1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+…+100) =2/(2*3)+2/(3*4)+2/(4*5)+…+2/(100*101) =2*[1/(2*3)+1/(3*4)+1/(4*5)+…+1/(100*101)] =2*(1/2-1/3+1/3-1/4+1/4-1/5+…+1/100-1/101) =2*(1/2-1/101) =99/101.

=1-1/2+1/2-1/3+1/3-1/4+...+1/2014-1/2015 =1-1/2015 =2014/2015

=1/2-1/3+1/3-1/4+1/4-1/5……+1/98-1/99+1/99-1/100 =1/2-1/100 =49/100 当然是我了

这是一个调和级数,根本没有什么简便方法,我们只有站在巨人的肩膀上来解这个题: Euler(欧拉)在1734年,利用Newton的成果,首先获得了调和级数有限多项和的值。 结果是:1+1/2+1/3+1/4+...+1/n= ln(n+1)+r【r的值,约为0.5772156649 称作的欧拉...

1-1/2+1/3-1/4+···+1/99-1/100 =1+1/2+1/3+1/4+···+1/99+1/100-2(1/2+1/4+···+1/100) =1+1/2+1/3+1/4+···+1/99+1/100-(1+1/2+···+1/50) =1/51+1/52+···+1/100

Private Sub Command1_Click() Dim S As Single Dim N As Integer N = Val(Text1) For i = 1 To N S = S + (-1) ^ (N - 1) * 1 / N Next Print "S = "; S End Sub

(-1又1/2)*(-1又1/3)*(-1又1/4)*····*(-1又1/2012) =-3/2*(-4/3)*(-5/4)*(-6/5)。。。。。*(-2011/2012)*(-2013/2012)(可以发现从第二项分母可以与第一项的分子约分) =(-1/2)*(-2013) =2013/2

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