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楼主 |
发表于 2025-5-21 06:55
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怎么来的?谢谢!!
\(\frac{2}{1}+\frac{6}{2}+\frac{12}{6}+\frac{20}{24}+\frac{30}{120}+\frac{42}{720}+\frac{56}{5040}\cdots+\frac{n(n+1)}{n!}=3e\)
源于这样一道题:
第1个问题。
2^1-1!=1,
1*2^2-2!=2,
2*2^3-3!=10,
10*2^4-4!=136,
136*2^5-5!=4232,
4232*2^6-6!=270128,
270128*2^7-7!=34571344,
34571344*2^8-8!=8850223744,
......
得到一串数——{1, 2, 10, 136, 4232, 270128, 34571344, 8850223744, 4531314194048, 4640065731076352, 9502854617204452096, 38923692512068956783616, 318862889058868887744361472,
5224249574340507856716440066048, 171188210051989761448883000409892864, 11218990533967201006313996293939948847104, 1470495527268148970299588122238941287859519488}——求通项公式——问题1。
第2个问题。
2^1-1=1,
1*2^2-3=1,
1*2^3-6=2,
2*2^4-10=22,
22*2^5-15=689,
689*2^6-21=44075,
44075*2^7-28=5641572,
5641572*2^8-36=1444242396,
1444242396*2^9-45=739452106707,
.......
得到一串数——{1, 1, 2, 22, 689, 44075, 5641572, 1444242396, 739452106707, 757198957267913, 1550743464484685758, 6351845230529272864690, 52034316128495803307540389,
852530235449275241390741733271, 27935710755201851109891825115824008, 1830794740052908514337870650790642188152, 239965928168214824791293381940431052885458791}——求通项公式——问题2。
第3个问题。
136/22=6.18182,
4232/689=6.14224,
270128/44075=6.12883,
34571344/5641572=6.12796,
......
5224249574340507856716440066048/852530235449275241390741733271=6.12793465511211873361212958209,
......
262345168553487441312403456944635391129575147770993115164671264739968851694392029493592064/
42811352163266089318903999189948042642370639369565874377218194446248920891354503995981524=6.12793465511211873361208276053,
......
这个6.12793465511211873361208276053, 最后=???——问题3——与前面3e的算式——有关系吗? |
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