摘要：又Sk+1=Sk+ak+1.所以+ak+1将ak=a1+(k-1)d代入上式.得(k+1)(a1+ak+1)=2ka1+k(k-1)d+2ak+1整理得(k-1)ak+1=(k-1)a1+k(k-1)d∵k≥2.∴ak+1=a1+［(k+1)-1］d.即n=k+1时等式成立.由①和②.等式对所有的自然数n成立.从而{an}是等差数列.

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