1.首项是a1,公差是d的等差数列的通项公式an=a1+(n-1)d,前n项和的公式是Sn=na1+(n(n-1))/2*d

当n=1时,S1=1*a1+(1*(1-1))/2d=a1
假设当n=k时,Sn=ka1+(k(k-1))/2*d
那么,当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k-1))/2*d+a1+(k+1-1)d
∵当n=k时,Sn=ka1+(k(k-1))/2*d
∴,当n=k+1时,Sn=ka1+(k(k-1))/2*d+a1+(k+1-1)d
,有Sn=(k+1)a1+((k+1)k)/2*d
.