[A.Maths] mathematical induction 好煩

Method 1:

RHS
= (k + 1)(2k2 + 7k + 3)
= k(2k2 + 7k + 3) + (2k2 + 7k + 3)
= 2k3 + 7k2 + 3k + 2k2 + 7k + 3
= 2k3 + 7k2 + 2k2 + 3k + 7k + 3
= 2k3 + 9k2 + 10k + 3
= LHS


Method 2:
LHS
= 2k3 + 9k2 + 10k + 3
= 2k3 + 7k2 + 3k + 2k2 + 7k + 3
= k(2k2 + 7k + 3) + (2k2 + 7k + 3)
= (k + 1)(k2 + 7k + 3)
= RHS

Method 3:
Let f(k) = 2k3 + 9k2 + 10k + 3
f(-1) = 2(-1)3 + 9(-1)2 + 10(-1) + 3 = 0
Hence, (k + 1) is one of the factor of f(k).

2k3 + 9k2 + 10k + 3 = (k + 1)(ak2 + bk + c)
Compare k3 terms: a = 2
Compare constant terms: c = 3
Compare k terms: 10 = b + c b = 7
Hence, 2k3 + 9k2 + 10k + 3 = (k + 1)(2k2 + 7k + 3)
=

我諗你同我一樣係f.4學生
我呀sir當教緊差唔多既題目果時 而我地又未學到fact 3次方 所以佢教我地咁做:
2k^3+9k^2+10k+3
=?
=(k+1)(2k^2+7k+3) ←抄答案

=2k^3+7k^2+3k+2k^2+7k+3 (再expend番佢)
↑之後就將佢放入"?"果度
咁就好似係自己factorize咁 不過其實係靠抄答案再expend 拎番個method mark

首先看最後的數是+3, 即是正1X正3=+3, 然後再將餘下的三個數除K, 這樣便可得出結果.