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1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7

拆分每个分数。记住等式: 1/(x(x+1))=1/x-1/(x+1) 所以,原等式就可以简化为1/1-1/10=0.9

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

1/2-1/3+1/4-1/5+1/6-1/7 =1/6+1/20+1/42 =7/42+1/42+1/20 =8/42+1/20 =80/420+21/420 =101/420

1/1*2*3+1/2*3*4+1/3*4*5+1/4*5*6+1/5*6*7+1/6*7*8+1/7*8*9+1/8*9*10+1/9*10*11 =(1/2)[(1/1*2)-(1/2*3)]+(1/2)[(1/2*3)-(1/3*4)]+(1/2)[(1/3*4)-(1/4*5)]+...+(1/2)[(1/9*10)-(1/10*11)] =(1/2)[(1/1*2)-(1/2*3)+(1/2*3)-(1/3*4)+(1/3*4)-(1/4*...

解法一: 1×2+2×3+3×4+...+n(n+1) =⅓×[1×2×3-0×1×2+2×3×4-1×2×3+3×4×5-2×3×4+...+n(n+1)(n+2)-(n-1)n(n+1)] =⅓n(n+1)(n+2) 解法二: 考察一般项第k项,k(k+1)=k²+k 1×2+2×3+3×4+...+n(n+1) =(1²+2²+3²+...+n...

#include int main() { float t=1; float m; int i,j; scanf("%d",&i); for(j=i;j>1;j--) { m=j*j; m=1/m; t=t-m; } printf("%f",t); } 我不知道你要哪一个的答案,我给你些了第二个,好像你的答案不对 第一个上面那位同志写的没有什么问题,我...

1/n(n+1)(n+2)(n+3) =1/(n^2+3n)(n^2+3n+2) =1/2[1/(n^2+3n)-1/(n^2+3n+2)] =1/2[1/n(n+3)-1/(n+1)(n+2)]...

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