Question

证明等差数列{an},(bn}的前n项和分别为Sn,Sn`,则am/bm=S2m-1/S2m`-1

Answer 1

证：

S(2m-1)/S'(2m-1)

=[a1+a2+...+a(2m-1)]/[b1+b2+...+b(2m-1)]

={[a1+a(2m-1)]+[a2+a(2m-2)]+...+[a(m-1)+a(m+1)]+am}/{[b1+b(2m-1)]+[b2+b(2m-2)]+...+[b(m-1)+b(m+1)]+bm}

=(2am+2am+...+2am+am)/(2bm+2bm+...+2bm+bm)

=(2m-1)am/[(2m-1)bm]

=am/bm