下面用数学归纳法证明Sn=na1+n(n-1)d/2和Sn=[a1(1-qⁿ)]/(1-q)（一）等差数列前n项和公式Sn=na1+n(n-1)d/2证明：(1)n=1,S1=a1,成立(2)设Sk=ka1+k(k-1)d/2,则S(k+1)=Sk+a(k+1)=ka1+k(k-1)d/2+a1+kd=(k+1)a1+(k+1...