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分数裂项求和方法总结(1) 用裂项法求1( 1)n n +型分数求和分析:因为1 11n n−+=1 1( 1) ( 1) ( 1)n nn n n n n n+− =+ + +(n 为自然数)所以有裂项公式:1 1 1( 1) 1n n n ...
1 1 1[ ]2 5 15115= −=(3) 用裂项法求 ( )kn n k+型分数求和分析:( )kn n k+型(n,k 均为自然数)1 1n n k−+=( ) ( )n k nn n k n n k+−+ +=( )kn n k+所...
1 1 1 1 1 1 1 1( ) ( ) ...... ( ) ( )1 3 3 5 3 5 5 7 93 95 95 97 95 97 97 991 11 3 97 9932009603= − + − + + − + −× × × × ×...
1 1 1 1 1 1( ) ( ) ...... ( )1 2 3 2 3 4 2 3 4 3 4 5 17 18 19 18 19 201 11 2 3 18 19 2011396840= − + − + + −× × × × × × × ...
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  1. 1. 分数裂项求和方法总结(1) 用裂项法求1( 1)n n +型分数求和分析:因为1 11n n−+=1 1( 1) ( 1) ( 1)n nn n n n n n+− =+ + +(n 为自然数)所以有裂项公式:1 1 1( 1) 1n n n n= −+ +【例1】 求1 1 1......10 11 11 12 59 60+ + +× × ×的和。1 1 1 1 1 1( ) ( ) ...... ( )10 11 11 12 59 601 110 60112= − + − + + −= −=(2) 用裂项法求1( )n n k+型分数求和分析:1( )n n k+型。(n,k 均为自然数)因为1 1 1 1 1( ) [ ]( ) ( ) ( )n k nk n n k k n n k n n k n n k+− = − =+ + + +所以1 1 1 1( )( )n n k k n n k= −+ +【例2】 计算1 1 1 1 15 7 7 9 9 11 11 13 13 15+ + + +× × × × ×1 1 1 1 1 1 1 1 1 1 1 1 1 1 1( ) ( ) ( ) ( ) ( )2 5 7 2 7 9 2 9 11 2 11 13 2 13 15= − + − + − + − + −1 1 1 1 1 1 1 1 1 1 1[( ) ( ) ( ) ( ) ( )]2 5 7 7 9 9 11 11 13 13 15= − + − + − + − + −1
  2. 2. 1 1 1[ ]2 5 15115= −=(3) 用裂项法求 ( )kn n k+型分数求和分析:( )kn n k+型(n,k 均为自然数)1 1n n k−+=( ) ( )n k nn n k n n k+−+ +=( )kn n k+所以( )kn n k+=1 1n n k−+【例3】 求2 2 2 2......1 3 3 5 5 7 97 99+ + + +× × × ×的和1 1 1 1 1 1 1(1 ) ( ) ( ) ...... ( )3 3 5 5 7 97 9911999899= − + − + − + + −= −=(4) 用裂项法求2( )( 2 )kn n k n k+ +型分数求和分析:2( )( 2 )kn n k n k+ +(n,k 均为自然数)2 1 1( )( 2 ) ( ) ( )( 2 )kn n k n k n n k n k n k= −+ + + + +【例4】 计算:4 4 4 4......1 3 5 3 5 7 93 95 97 95 97 99+ + + +× × × × × × × ×2
  3. 3. 1 1 1 1 1 1 1 1( ) ( ) ...... ( ) ( )1 3 3 5 3 5 5 7 93 95 95 97 95 97 97 991 11 3 97 9932009603= − + − + + − + −× × × × × × × ×= −× ×=(5) 用裂项法求1( )( 2 )( 3 )n n k n k n k+ + +型分数求和分析:1( )( 2 )( 3 )n n k n k n k+ + +(n,k 均为自然数)1 1 1 1( )( )( 2 )( 3 ) 3 ( )( 2 ) ( )( 2 )( 3 )n n k n k n k k n n k n k n k n k n k= −+ + + + + + + +【例5】 计算:1 1 1......1 2 3 4 2 3 4 5 17 18 19 20+ + +× × × × × × × × ×1 1 1 1 1 1 1[( ) ( ) ...... ( )]3 1 2 3 2 3 4 2 3 4 3 4 5 17 18 19 18 19 201 1 1[ ]3 1 2 3 18 19 20113920520= − + − + + −× × × × × × × × × × × ×= −−× × × ×=(6) 用裂项法求3( )( 2 )( 3 )kn n k n k n k+ + + 型分数求和分析:3( )( 2 )( 3 )kn n k n k n k+ + +(n,k 均为自然数)3 1 1( )( 2 )( 3 ) ( )( 2 ) ( )( 2 )( 3 )kn n k n k n k n n k n k n k n k n k= −+ + + + + + + +【例6】 计算:3 3 3......1 2 3 4 2 3 4 5 17 18 19 20+ + +× × × × × × × × ×3
  4. 4. 1 1 1 1 1 1( ) ( ) ...... ( )1 2 3 2 3 4 2 3 4 3 4 5 17 18 19 18 19 201 11 2 3 18 19 2011396840= − + − + + −× × × × × × × × × × × ×= − −× × × ×=(七)用裂项法求复合型分数和(例题略)4

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