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1/1*2+1/2*3+1/3*4······+1/2013*2014

(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...

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

1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+···+1/99*100 =(1-1/2)+(1/2-1/3)+(1/3-1/4)...+(1/99-1/100) =1-1/2+1/2-1/3+1/3-1/4+....1/99-1/100 =1-1/100 =99/100

(-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

你好,程序如下: #include void main() { int i; double sum = 0.0; for(i = 1;i < 16;i ++) { sum += 1.0/i; } printf("%lf\n",sum); } 有疑问的话欢迎交流。

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.

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

调和数列,没有公式。只能用Excel或编程计算。 1-1/2+1/3-1/4+···+1/99-1/100≈0.68817

=1÷1/2÷2/3÷……÷2013/2014 =1×2×3/2×4/3×……×2014/2013 =1×2014(把分母都约下去) =2014 望采纳

=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

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