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.