Through formula calculation, it is concluded that S 1=a 1, so the proposition holds when n= 1.

Step 2: If n=k, Sn = a1(1-qk)/(1-q).

When n=k+ 1, SK+ 1 plus ak+ 1(k+ 1 is subscript) = A 1 (1-q k)/(65438).

+a 1*q^k

= (a11-q) (1-qk+qk-q (k+1)) (this step is to extract the common factor) = a 1 (1-)

So the proposition holds when n=k+ 1

From the first step and the second part, we know that for any n(n is a positive integer), Sn = a1(1-q n)/(1-q).