李毓佩《数学历险记》---裂项法简算分数

2023-12-29 22:20:08

李毓佩《数学历险记》—裂项法简算分数

问题描述：

计算: 1 1 ? 2 + 1 2 ? 3 + 1 3 ? ４ ＋　．．．＋ 1 98 ? 99 + 1 99 ? 100 \frac{1}{1 * 2} + \frac{1}{2 * 3}+ \frac{1}{3 *４}＋　．．．＋\frac{1}{98 * 99} + \frac{1}{99 * 100} 1?21?+2?31?+3?４1?＋　．．．＋98?991?+99?1001?
1 1 ? 2 + 1 2 ? 3 + 1 3 ? 4 ＋　．．．＋ 1 98 ? 99 + 1 99 ? 100 \frac{1}{1 * 2} + \frac{1}{2 * 3}+ \frac{1}{3 *4}＋　．．．＋\frac{1}{98 * 99} + \frac{1}{99 * 100} 1?21?+2?31?+3?41?＋　．．．＋98?991?+99?1001?
= ( 1 1 ? 1 2 ) + ( 1 2 ? 1 3 ) + ( 1 3 ? 1 4 ) + . . . . . + ( 1 98 ? 1 99 ) + ( 1 99 ? 1 100 ) = (\frac{1}{1} - \frac{1}{2}) +(\frac{1}{2} - \frac{1}{3}) +(\frac{1}{3} - \frac{1}{4}) +.....+(\frac{1}{98} - \frac{1}{99})+(\frac{1}{99} - \frac{1}{100}) =(11??21?)+(21??31?)+(31??41?)+.....+(981??991?)+(991??1001?)
= 1 ? 1 100 =1 - \frac{1}{100} =1?1001?
= 99 100 =\frac{99}{100} =10099?
把一个分数拆成两个分数相减的形式，就是裂项法，可以根据裂项工式: 1 a ? b = ( 1 a ? 1 b ) ? 1 b ? a \frac{1}{a * b} = (\frac{1}{a} - \frac{1}{b}) * \frac{1}{b - a} a?b1?=(a1??b1?)?b?a1?把一个数裂项为两个分数求差，然后前后抵消求和。裂项法是分解与组合思想在数列求和中的具体应用，通常用于代数，分数，有时候也用于整数数列求和。

裂相法求各工式
  1. 1 n ( n + 1 ) = 1 n ? 1 n + 1 \frac{1}{n(n +1)} = \frac{1}{n} - \frac{1}{n+1} n(n+1)1?=n1??n+11?
  2. 1 ( 2 n ? 1 ) ( 2 n + 1 ) = 1 2 ( 1 2 n ? 1 ? 1 2 n + 1 ) \frac{1}{(2n-1)(2n +1)} = \frac{1}{2}(\frac{1}{2n-1} - \frac{1}{2n+1}) (2n?1)(2n+1)1?=21?(2n?11??2n+11?)
  3. 1 n ( n + 1 ) ( n + 2 ) = 1 2 ( 1 n ( n ? 1 ) ? 1 ( n + 1 ) ( n + 2 ) ) \frac{1}{n(n+1)(n +2)} = \frac{1}{2}(\frac{1}{n(n-1)} - \frac{1}{(n+1)(n+2)}) n(n+1)(n+2)1?=21?(n(n?1)1??(n+1)(n+2)1?)
  4. 1 a + b = 1 a ? b ( a ? b ) \frac{1}{\sqrt{a}+\sqrt{b}} = \frac{1}{a - b} (\sqrt{a} - \sqrt{b}) a ?+b ?1?=a?b1?(a ??b ?)
  5. n ? n ！ = ( n + 1 ) ! ? n ! n * n！ = (n + 1)! - n! n?n！=(n+1)!?n!
  6. 1 n ( n + k ) = 1 k ( 1 n ? 1 n + k ) \frac{1}{n(n+k)} = \frac{1}{k}(\frac{1}{n} - \frac{1}{n +k}) n(n+k)1?=k1?(n1??n+k1?)
  7. 1 n + n + 1 = n + 1 ? n \frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1} - \sqrt{n} n ?+n+1 ?1?=n+1 ??n ?
  8. 1 n + n + k = 1 k ( n + k ? n ) \frac{1}{\sqrt{n}+\sqrt{n+k}} = \frac{1}{k} (\sqrt{n+k} - \sqrt{n}) n ?+n+k ?1?=k1?(n+k ??n ?)
例: 1 1 ? 2 ? 3 + 1 2 ? 3 ? 4 + 1 3 ? 4 ? 5 ＋ . . . + 1 97 ? 98 ? 99 + 1 98 ? 99 ? 100 \frac{1}{1 * 2*3} + \frac{1}{2*3*4}+\frac{1}{3 *4*5}＋...+\frac{1}{97*98 * 99} + \frac{1}{98*99 * 100} 1?2?31?+2?3?41?+3?4?51?＋...+97?98?991?+98?99?1001?

= 1 2 ( 1 1 ? 2 ? 1 2 ? 3 + 1 2 ? 3 ? 1 3 ? 4 + 1 3 ? 4 ? 1 4 ? 5 + . . . . . + 1 97 ? 98 ? 1 98 ? 99 + 1 98 ? 99 ? 1 99 ? 100 ) =\frac{1}{2}(\frac{1}{1*2} - \frac{1}{2*3} +\frac{1}{2*3} - \frac{1}{3*4} +\frac{1}{3*4} - \frac{1}{4*5} +.....+\frac{1}{97*98} - \frac{1}{98*99}+\frac{1}{98*99} - \frac{1}{99*100}) =21?(1?21??2?31?+2?31??3?41?+3?41??4?51?+.....+97?981??98?991?+98?991??99?1001?)
= 1 2 ( 1 1 ? 2 ? 1 99 ? 100 ) =\frac{1}{2}(\frac{1}{1*2} - \frac{1}{99*100}) =21?(1?21??99?1001?)
= 4949 19800 =\frac{4949}{19800} =198004949?

文章来源:https://blog.csdn.net/qq_31165949/article/details/135187250
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