2016年12月29日木曜日

学習環境

数学読本〈3〉平面上のベクトル/複素数と数列とその和/空間図形/2次曲線/数列(松坂 和夫(著)、岩波書店)の第13章(「離散的」な世界 - 数列)、13.1(数列とその和)、平方の和、立方の和、問20、21.を取り組んでみる。

問20.


  1. 3 n( n+1 ) 2 2n= n( 3n1 ) 2

  2. i=1 n ( 2 i 2 +i1 ) = 2n( 2n+1 )( n+1 ) 6 + n( n+1 ) 2 n = n( 4 n 2 +6n+2+3n+36 ) 6 = n( 4 n 2 +9n1 ) 6

  3. i=1 n ( i 2 +2i ) = n( 2n+1 )( n+1 ) 6 +n( n+1 ) = n( n+1 )( 2n+7 ) 6

  4. n( 2n+1 )( n+1 ) 6 + 3n( n+1 ) 2 4n = n( 2 n 2 +3n+1+9n+924 ) 6 = n( 2 n 2 +12n14 ) 6 = n( n 2 +6n7 ) 3 = n( n+7 )( n1 ) 3
  5. p

  6. ( n( n+1 ) 2 ) 2 + 2n( 2n+1 )( n+1 ) 6 +n = n 2 ( n+1 ) 2 4 + 2n( 2n+1 )( n+1 ) 6 + n( n+1 ) 2 = n( n+1 )( 3n( n+1 )+4( 2n+1 )+6 ) 12 = n( n+1 )( 3 n 2 +3n+8n+4+6 ) 12 = n( n+1 )( 3 n 2 +11n+10 ) 12 = n( n+1 )( n+2 )( 3n+5 ) 12

  7. 2 n 2 ( n+1 ) 2 4 + 3n( 2n+1 )( n+1 ) 6 + n( n+1 ) 2 = n( n+1 )( n( n+1 )+( 2n+1 )+1 ) 2 = n( n+1 )( n 2 +3n+2 ) 4 = n ( n+1 ) 2 ( n+2 ) 4

問21.


  1. a n = ( 2n1 ) 2 =4 n 2 4n+1 4n( 2n+1 )( n+1 ) 6 4n( n+1 ) 2 +n = n( 4 n 2 +6n+26n6+3 ) 3 = n( 4 n 2 1 ) 3 = n( 2n+1 )( 2n1 ) 3

  2. a n = ( 2n ) 2 =4 n 2 4n( 2n+1 )( n+1 ) 6 = 2n( 2n+1 )( n+1 ) 3

  3. a n = n 2 ( 3n1 )=3 n 3 n 2 3 n 2 ( n+1 ) 2 4 n( 2n+1 )( n+1 ) 6 = n( n+1 )( 9n( n+1 )2( 2n+1 ) ) 12 = n( n+1 )( 9 n 2 +9n4n2 ) 12 = n( n+1 )( 9 n 2 +5n2 ) 12

  4. a n = n( n+1 ) 2 = 1 2 ( n 2 +n ) 1 2 ( n( 2n+1 )( n+1 ) 6 + n( n+1 ) 2 ) = 2n( n+1 )( n+2 ) 12 = n( n+1 )( n+2 ) 6

  5. n( n+1 ) 2 ( n( n+1 ) 2 +1 ) 2 = 1 8 n( n+1 )( n( n+1 )+2 ) = n( n+1 )( n 2 +n+2 ) 8

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