YLL討論網

對於每一個自然數n, 證明:




此題為:
1990年環球城市數學競賽春季賽國中組初級卷第1題(4分)

用數學歸納法
當n=1時
命題成立

(1+1/2+...1/k+1/(k+1))^2+(1/2+...+1/k+1/(k+1))^2...+(1/k+1/(k+1))^2+(1/(k+1))^2
=(1+1/2...1/k)^2+2(1+1/2+1/k)(1/(k+1))+(1/(k+1))^2+(1/2...+1/k)^2+2(1/2+...1/k)(1/(k+1))+(1/(k+1))^2...+1/k^2+2(1/k)(1/(k+1))+(1/(k+1))^2
=[(1+1/2+1/k)^2+(1/2...+1/(k+1))^2+...+(1/k)^2]+2(1/(k+1))([1+1/2+...1/k]+[1/2...+1/k]+...+1/k)+(k+1)(1/(k+1))^2
=2k-(1+1/2+1/3+...1/k)+2(1/(k+1))(1+1+1...+1{k個1})+1/(k+1)
=2k-(1+1/2+1/3+...1/k)+(2k+1)(1/(k+1))
=2k-(1+1/2+1/3+...1/k)+2(k+1)(1/(k+1))-1/(k+1)
=2(k+1)-(1+1/2+1/3+...1/k+1/(k+1))
所以n=任何自然數，命題也成立


PS好似寫得很亂...
自己也有些看不懂...
有錯別見怪...