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熱心網友

證明：設等差數列{an}的公差為d左邊=[(√a2-√a1)/(a2-a1)]+[(√a3-√a2)/(a3-a2)]+...+[(√an-√an_1)/(an-an_1)]=[(√an-√a1)/d]=(an-a1)/[(√an+√a1)d]=[(n-1)d]/[(√an+√a1)d]=(n-1)/(√an+√a1)=右邊證畢