The general formula of geometric series: an= a 1 qn- 1.

1, a (1) = a, and a (n) is the arithmetic progression with error r.

1- 1, general formula,

a(n)=a(n- 1)+r=a(n-2)+2r=...= a[n-(n- 1)]+(n- 1)r = a( 1)+(n- 1)r = a+(n- 1)r。

It can be proved by induction.

When n= 1, a (1) = a+(1-1) r = a.

Assuming n=k, arithmetic progression's general formula holds. a(k)=a+(k- 1)r

Then, when n=k+ 1, a (k+1) = a (k)+r = a+(k-1) r+r = a+[(k+1)-/kl.

The general formula also holds.

Therefore, through induction, arithmetic progression's general formula is correct.

1-2, summation formula,

S(n)=a( 1)+a(2)+...+a(n)

=a+(a+r)+...+[a+(n- 1)r]

=na+r[ 1+2+。 ...+(n- 1)]

=na+n(n- 1)r/2

Similarly, the summation formula can also be proved by induction. (omitted)

2, a (1) = a, and a (n) is the geometric series of the common ratio r(r is not equal to 0).

2- 1, general term formula,

a(n)=a(n- 1)r=a(n-2)r^2=...=a[n-(n- 1)]r^(n- 1)=a( 1)r^(n- 1)=ar^(n- 1).

The general formula of geometric series can be proved by induction. (omitted)

2-2, summation formula,

S(n)=a( 1)+a(2)+...+a(n)

=a+ar+...+ar^(n- 1)

= a[ 1+r+0...+r^(n- 1)]

When r is not equal to 1,

s(n)=a[ 1-r^n]/[ 1-r]

When r= 1,

S(n)=na。

Similarly, the summation formula can also be proved by induction.