| (i)n=1のとき (左辺)=1-1/2=1/2 (右辺)=1/(1+1)=1/2 となり成立。
(ii)n=kのとき成立を仮定します。 つまり 1-1/2+…+1/(2k-1)-1/(2k)=1/(k+1)+…1/(2k) (A) n=k+1のとき (左辺)=1-1/2+…+1/(2k-1)-1/(2k)+1/(2k+1)-1/(2k+2) ={1/(k+1)+…1/(2k)}+1/(2k+1)-1/(2k+2) (∵)(A)を代入 =1/(k+2)+…1/(2k)+1/(2k+1)-1/(2k+2)+1/(k+1) =1/{(k+1)+1}+…1/(2k)+1/(2k+1)-(1/2)/(k+1)+1/(k+1) =1/{(k+1)+1}+…1/(2k)+1/(2k+1)+(1/2)/(k+1) =1/{(k+1)+1}+…1/(2k)+1/(2k+1)+1/{2(k+1)} =(右辺) となりこのときも成立。
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