Original formula = (a+a 2+a 3+...+a n)-(1+2+3+...+n)

When a= 1,

=n-[n(n+ 1)/2]

=-n^2/2+n/2

When a is not equal to 1,

=a( 1-a^n)/( 1-a)-n(n+ 1)/2

2) and (2-3 * 5-1)+(4-3 * 5-2)+...+(2n-3 * 5-n)

Original formula = (2+4+6+...+2n)-(3 * 5 (-1)+3 * 5 (-2)+...+3 * 5 (-n))

=(2+2n)n/2-3/5( 1- 1/5^n)/( 1- 1/5)

=n(n+ 1)-3( 1- 1/5^n)/4

Ps: Memorize the summation formula of arithmetic progression and proportional sequence.

The condition of the summation formula q of equal proportion sequence.