✔ 最佳答案
1)A1 = (1 + 3/1) = 4A2 = (1 + 3/1)(1 + 5/4) = 9A3 = (1 + 3/1)(1 + 5/4)(1 + (2*3+1)/3^2) = 162)推測 An = (n+1)^2設 (1+3/1)(1+5/4)...(1+(2n+1)/n^2) = (n+1)^2 對正整數 k 成立,即 Ak = (1+3/1)(1+5/4)...(1+(2k+1)/k^2) = (k+1)^2由1) 已驗證 A1 為真,當 n = k+1 ,A(k+1) =(1+3/1)(1+5/4)...(1+(2k+1)/k^2) * [1 + (2(k+1)+1)/(k+1)^2]= (k+1)^2 * [1 + (2(k+1)+1)/(k+1)^2]= (k+1)^2 * [1 + (2k+3)/(k+1)^2]= (k+1)^2 + (2k+3)= k^2 + 2k + 1 + 2k + 3= k^2 + 4k + 4= (k + 2)^2當 n = k 為真 , n = k+1 亦真 , 由數學歸納法得證。
2010-09-06 21:14:12 補充：
1)
A1 = (1 + 3/1) = 4
A2 = (1 + 3/1)(1 + 5/4) = 4(9/4) = 9
A3 = (1 + 3/1)(1 + 5/4)(1 + (2*3+1)/3^2)
= 9(1 + (2*3+1)/3^2)
= 9(1 + 7/9)
= 9(16/9)
= 16
2010-09-06 21:16:32 補充：
A1 得一項 , n=1 ,
代入 (1+(2n+1)/n^2) 得
= (1 + 3/1)
2010-09-06 21:21:31 補充：
A2 有二項 , n=2 ,
代入(1+(2n+1)/n^2) 得 (1 + 5/4) ,
要乘番 A1 的項才有二項 ,
這是題目An = (2+3/1)(1+5/4)...(1+(2n+1)/n^2) 所定義的 ,
An 有 n 項相乘。
2010-09-06 21:25:26 補充：
An = (1+3/1)(1+5/4)...(1+(2n+1)/n^2)
你打錯了(2+3/1)(1+5/4)...(1+(2n+1)/n^2)
