Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1) 02/08/2021 Bởi Katherine Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
3A = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 +… + n . ( n + 1) . 3 = 1 . 2 . ( 3 – 0 ) + 2 . 3 . ( 4 – 1 ) + … + n . ( n + 1 ) .[ ( n + 2 ) – ( n – 1 ) ] =[ 1 . 2 . 3 + 2 . 3 . 4 + …+ ( n – 1 ) . n . ( n + 1 ) + n . ( n + 1) . ( n + 2 ) ] – [ 0 . 1 . 2 + 1 . 2 . 3 +…+ ( n – 1 ) . n . ( n + 1 ) ] = n . ( n + 1 ) . ( n + 2 ) =>A = [ n . ( n + 1 ) . ( n + 2 ) ] /3 Bình luận
`A = 1.2 + 2.3 + 3.4 + … + n(n+1)` `=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n+1).3 ` `= 1.2.3 + 2.3(4-1) + 3.4(5-1) + … + n(n+1)(n+2 – n-1)` `=> 3A – A = 2A = 1.2.3 + 2.3.4 – 1.2.3 + 3.4.5 – 2.3.4 + … + n(n+1)(n+2) – (n-1)n(n+1)` `=> A = \frac{n(n+1)(n+2)}{2}` Bình luận
3A = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 +… + n . ( n + 1) . 3
= 1 . 2 . ( 3 – 0 ) + 2 . 3 . ( 4 – 1 ) + … + n . ( n + 1 ) .[ ( n + 2 ) – ( n – 1 ) ]
=[ 1 . 2 . 3 + 2 . 3 . 4 + …+ ( n – 1 ) . n . ( n + 1 ) + n . ( n + 1) . ( n + 2 ) ] – [ 0 . 1 . 2 + 1 . 2 . 3 +…+ ( n – 1 ) . n . ( n + 1 ) ]
= n . ( n + 1 ) . ( n + 2 )
=>A = [ n . ( n + 1 ) . ( n + 2 ) ] /3
`A = 1.2 + 2.3 + 3.4 + … + n(n+1)`
`=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n+1).3 `
`= 1.2.3 + 2.3(4-1) + 3.4(5-1) + … + n(n+1)(n+2 – n-1)`
`=> 3A – A = 2A = 1.2.3 + 2.3.4 – 1.2.3 + 3.4.5 – 2.3.4 + … + n(n+1)(n+2) – (n-1)n(n+1)`
`=> A = \frac{n(n+1)(n+2)}{2}`