(1).当n＝1时,左边＝a1^2,右边＝a1^2,命题成立.
(2).假设当n＝k时命题成立,即
:(a1+a2+…+ak)^2=a1^2+a2^2+…ak^2+2[a1a2+a1a3+…a(k-1)ak].…………………①
那么．当n＝k＋1时,
〔(a1＋a2＋…＋ak)＋a(k+1)〕^2
＝(a1＋a2＋…＋ak)＾2＋a(k+1)^2＋2a(k+1)(a1＋a2＋… ＋ak)…………………②
把①式代入②式得．
〔(a1＋a2＋…＋ak)＋a(k+1)〕^2
＝a1^2+a2^2+…ak^2+2[a1a2+a1a3+…a(k-1)ak]＋a(k+1)^2 ＋2a(k+1)(a1＋a2＋…＋ak)
＝〔a1^2+a2^2+…ak^2＋a(n+1)^2]+2[a1a2+a1a3+…a(k-1)ak+aka(k+1)]
所以,当n＝k+1时命题也成立．
综合上述可知,(a1+a2+…+an)^2=a1^2+a2^2+…an^2+2(a1a2+a1a3+…an-1an)对任意非零自然数n都成立．