Amath問題(M.I.)

1 + 2 + ... + n is an A.P. where the 1st term is 1 and common difference is 1, so
T ( n ) = a + ( n - 1 )( d )
= 1 + ( n - 1 )( 1 )
= n
Then S ( n ) = ( 1st term + last term )( no. of terms ) / 2
= ( 1 + n )( n ) / 2
Referring to your question,
1 x 2 + 2 x 3 +...+ n (n + 1 )
=1 x ( 1 + 1 ) + 2 ( 2 + 1 ) + ... + n ( n + 1 )
= ( 1^2 + 2^2 + ... + n^2 ) + ( 1 + 2 + ... + n )
= n ( n + 1 )( 2n + 1 ) / 6 + n ( n + 1 ) / 2
= n ( n + 1 )( n + 2 ) / 3


2008-01-22 22:23:07 補充：
Thank you for inviting me to answer this question.

[n(n+1)]/2是連續數之和的公式=1+2+3+...+n
[項數(n)]*([尾項(n)]+[頸項(1)])/2
([尾項(n)]+[頸項(1)])/2=中間數

2008-01-25 12:15:07 補充：
不是(頸)項, 應該是首項