a(n) = a + (n-1)d.
用數(shù)學歸納法證明,s(n) = na + n(n-1)d/2.
n=1時,s(1) = a(1) = a = 1*a + 1*(1-1)d/2,滿足題意.
設(shè)n=k時,有s(n) = na + n(n-1)d/2.
則有,s(k) = ka + k(k-1)d/2.
當n=k+1時,s(k+1) = s(k) + a(k+1) = ka + k(k-1)d/2 + [a + kd] = (k+1)a + kd[(k-1)/2 + 1]
= (k+1)a + kd(k+1)/2
= (k+1)a + (k+1)(k+1-1)d/2,
也滿足題意.
因此,由歸納法知,總有,s(n) = na + n(n-1)d/2.
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樓主問題中,第一個等號說的是,當n=k+1時,要證明的結(jié)論；
第二個等號說的是,s(k+1) = s(k) + a(k+1).
俺說清楚點沒.那請問為什么s(k+1) = s(k) + a(k+1)s(k) = a(1)+a(2)+...+a(k).
s(k+1)=a(1)+a(2)+...+a(k)+a(k+1) = [a(1)+a(2)+...+a(k)] + a(k+1) = s(k) + a(k+1)