等差数列公式证明：
(1)n=1,S1=a1,成立
(2)设Sk=ka1+(1/2)k(k-1)d,则Sk+1=Sk+ak+1=ka1+(1/2)k(k-1)d+a1+kd
=(k+1)a1+(1/2)(k+1)kd,所以n=k+1也成立.
等比数列
(1)n=1,S1=a1成立
(2)Sk+1=Sk+ak+1=a1(1-q^k)/(1-q)+a1q^k
=[a1/(1-q)][1-q^k+q^k-q^(k+1)]
=a1[1-q^(k+1)]/(1-q)
所以n=k+1时公式仍成立.
综上,两个公式都成立.