Sn=n*a1+n(n-1)d/2当n=1时S1=a1.成立假设n=k时,有Sk=k*a1+k(k-1)d/2等式成立则有n=k+1时S（k+1)=Sk+a(k+1)=Sk+a1+k*d=k*a1+k(k-1)d/2+a1+k*d=(k+1)*a1+((k²-k)d+2k*d)/2=(k+1)*a1+(k²+k)d/2=(k+1)*a1+(k+1...