數學歸納法

case for n = 1 簡單, 自己寫
設 3/4+5/36+7/144+...+(2k+1)/k²(k+1)² = 1- 1/(k+1)²

3/4+5/36+7/144+...+ (2k+1)/k²(k+1)² + (2k+3)/(k+1)²(k+2)²
= 1- 1/(k+1)² + (2k+3)/(k+1)²(k+2)²
= 1 - [ (k+2)² - 2k - 3 ] / [ (k+1)²(k+2)² ]
= 1 - [ k² + 4k + 4 - 2k - 3 ] / [ (k+1)²(k+2)² ]
= 1 - [ k² + 2k + 1 ] / [ (k+1)²(k+2)² ]
= 1 - 1/(k+2)²


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(1)(1+1)(2+1) = 6 被3整除

設 k(k+1)(2k+1) = 3M, M 為整數
(k+1)(2k^2 + k) = 3M

(k+1)(k+2) [2(k+1)+1]
= (k+1)(k+2)(2k+3)
= (k+1)(2k^2 + 7k + 6)
= (k+1)(2k^2 + k) + (k+1)(6k + 6)
= 3M + 3(k+1)(2k + 2)
能被3整除

自己寫埋 d 公式野....