S1=a1+0d=a1成立
假设n=k时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)(k+1-1)d/2也成立,
综上,等差数列的前n项和公式Sn=na1+n(n-1)d/2