Question

Tính A=1.2+2.3+3.4+4.5+.........+n.(n+1)

Answer 1

Ta có : 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + n.( n + 1 ).3

=> 3A = 1.2.( 3 - 0 ) + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ..... + n.( n + 1 ).[ ( n + 2 ) - ( n - 1 ) ]

=> 3A = 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 )

=> 3A = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + [ ( n - 1 ).n.( n + 1 ) - ( n - 1 ).n.( n + 1 ) ] + n.( n + 1 ).( n + 2 )

=> 3A = n.( n + 1 ).( n + 2 )

=> A = \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)