昨天睡的早，想着今天回，却找不到了。
1/[n(n+1)(n+2)(n+3)] = {[1/(n(n+1)) + 1/((n+2)(n+3))] / 2 - 1/[(n+1)(n+2)] } / 3
每一项恰好是两子部分的和的一半，减去中间那部分，从直觉上讲，正好会“消去”
每项的中间那部分，在前一项和后一项是加1/2个的，式子中是减的，所以能消去
化简
前n项和为:
sum(n) = 1/18 - 1/[3(n+1)(n+2)(n+3)]
写个代码验证之：
#include <stdio.h>
int main()
{
double sum1, sum2;
int i, j;
for (i=1; i<100; i++) {
sum1 = 0.0;
for (j=1; j<=i; j++)
sum1 += 1.0/(j*(j+1)*(j+2)*(j+3));
sum2 = 1.0/18 - 1.0/(3*(i+1)*(i+2)*(i+3));
printf("%dth %f %f\n", i, sum1, sum2);
}
}
输出：
1th 0.041667 0.041667
2th 0.050000 0.050000
3th 0.052778 0.052778
4th 0.053968 0.053968
5th 0.054563 0.054563
6th 0.054894 0.054894
7th 0.055093 0.055093
8th 0.055219 0.055219
9th 0.055303 0.055303
10th 0.055361 0.055361
11th 0.055403 0.055403
12th 0.055433 0.055433
13th 0.055456 0.055456
14th 0.055474 0.055474
15th 0.055487 0.055487
16th 0.055498 0.055498
17th 0.055507 0.055507
18th 0.055514 0.055514
19th 0.055519 0.055519
20th 0.055524 0.055524
21th 0.055528 0.055528
22th 0.055531 0.055531
23th 0.055534 0.055534
24th 0.055537 0.055537
25th 0.055539 0.055539
26th 0.055540 0.055540
27th 0.055542 0.055542
28th 0.055543 0.055543
29th 0.055544 0.055544
30th 0.055545 0.055545
31th 0.055546 0.055546
32th 0.055547 0.055547
33th 0.055548 0.055548
34th 0.055548 0.055548
35th 0.055549 0.055549
36th 0.055549 0.055549
37th 0.055550 0.055550
38th 0.055550 0.055550
39th 0.055551 0.055551
40th 0.055551 0.055551
41th 0.055551 0.055551
42th 0.055552 0.055552
43th 0.055552 0.055552
44th 0.055552 0.055552
45th 0.055552 0.055552
46th 0.055553 0.055553
47th 0.055553 0.055553
48th 0.055553 0.055553
49th 0.055553 0.055553
50th 0.055553 0.055553
51th 0.055553 0.055553
52th 0.055553 0.055553
53th 0.055554 0.055554
54th 0.055554 0.055554
55th 0.055554 0.055554
56th 0.055554 0.055554
57th 0.055554 0.055554
58th 0.055554 0.055554
59th 0.055554 0.055554
60th 0.055554 0.055554
61th 0.055554 0.055554
62th 0.055554 0.055554
63th 0.055554 0.055554
64th 0.055554 0.055554
65th 0.055554 0.055554
66th 0.055554 0.055554
67th 0.055555 0.055555
68th 0.055555 0.055555
69th 0.055555 0.055555
70th 0.055555 0.055555
71th 0.055555 0.055555
72th 0.055555 0.055555
73th 0.055555 0.055555
74th 0.055555 0.055555
75th 0.055555 0.055555
76th 0.055555 0.055555
77th 0.055555 0.055555
78th 0.055555 0.055555
79th 0.055555 0.055555
80th 0.055555 0.055555
81th 0.055555 0.055555
82th 0.055555 0.055555
83th 0.055555 0.055555
84th 0.055555 0.055555
85th 0.055555 0.055555
86th 0.055555 0.055555
87th 0.055555 0.055555
88th 0.055555 0.055555
89th 0.055555 0.055555
90th 0.055555 0.055555
91th 0.055555 0.055555
92th 0.055555 0.055555
93th 0.055555 0.055555
94th 0.055555 0.055555
95th 0.055555 0.055555
96th 0.055555 0.055555
97th 0.055555 0.055555
98th 0.055555 0.055555
99th 0.055555 0.055555
