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2 4 3 9 5 20 7

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

这种多项分数加减的题目,多数是用所谓的列项法来做。 3/2-5/6+7/12-9/20+11/30-13/42 =1+1/2-(1/2+1/3)+(1/3+1/4)-(1/4+1/5)+(1/5+1/6)-(1/6+1/7) =1+1/2-1/2-1/3+1/3+1/4-1/4-1/5+1/5+1/6-1/6-1/7 =1-1/7 =6/7 裂项法的实质是将数列中的每项(...

3/4+2/5+7/8+9/20+12/35+13/40 =3/4+7/8+2/5+12/35+9/20+13/40 =13/8十26/35+31/40 =13/8+31/40+26/35 =96/40十26/35 =12/5+26/35 =110/35 =22/7

#include int main() {int i,j; printf("* "); for(i=1;i

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

采纳哦

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#include main() { long int i,sum=0,temp=1;/* 定义为长整型 */ for(i=1;i

3/2-5/6+7/12-9/20+11/30-……+99/2450 =(1+1/2)-(1/2+1/3)+(1/3+1/4)-(1/4+1/5)+(1/5+1/6)-......+(1/49+1/50) =1+1/2-1/2-1/3+1/3+1/4-1/4-1/5+1/5+1/6-......+1/49+1/50 =1+1/50 =51/50

原式=3/2-(1/2+1/3)+(1/3+1/4)-(1/4+1/5)+(1/5+1/6)-(1/6+1/7)+(1/7+1/8) =3/2-1/2-1/3+1/3+1/4-1/4-1/5+1/5+1/6-1/6-1/7+1/7+1/8 =3/2-1/2+1/8 =1+1/8 =9/8

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