你第3個= 係要自己夾硬加左-2k-3 係個括號到 你要係後面+返6k +9 先抵銷前邊 而個-2k-5 係本身有

留意括號內嗰堆嘢，就係要「鬥」一舊 f(k) 出嚟，咁「鬥」完舊嘢就梗係要扣返啲殘渣啦！但整體都要符合返上個 step 先得㗎！ 唔好忘記已經假設咗 f(k)=4t。

6k + 9 -2k - 5 is not actually constructed instead of calculated. In order to prove the formula works when k=k+1, you need to make use of the previous result when when n=kn is true. Since the previous step have already proven f(k) = 3^(k+1) -2k - 3 is true. What you need to do is to construct the first half of the calculation to look EXACTLY as 3^(k+1) -2k - 3 .

抵消返多左既野出黎

It is very common to reuse previous result when you perform calculation. When you deal with calculation or integration properties (most usually involved with trigonometric functions), you will most likely need to applies the result from the previous step.

But actually the calculation is a bit foolish in my view. When you reach to f(k+1)!= 3(4t) + 4k + 4, you can simply rewrite it as f(k+1) = 4(3t + k + 1) = 4x where x is any positive integer.

矩跳左個stEp 本身係 3*2k+3

技巧黎，之後你會越黎越多+a-a 乘a除a，之類，睇黎你要操多啲唔同題目

-2=3(-2)+6-2

與其咁樣寫，不如將3^(k+1)寫做4t+2k+3