Example 1
(a) Prove the product of three consecutive integers is divisible by \( 3 \).
\( \begin{align} &\text{Consider three consecutive positive integers are } n,n+1,n+2 \text{ where } n \text{ is a positive integer} \\ &\text{So the product of three consecutive positive integers: } n(n+1)(n+2) \\ &n \text{ can be in the form of } 3k, 3k+1 \text{ or } 3k+2 \text{ where } k \text{ is a positive integer} \end{align} \)
\( \require{AMSsymbols} \begin{align} \text{For } n &= 3k, \\ n(n+1)(n+2) &= 3k(3k+1)(3k+2) \\ &= \color{red}3 \times k(3k+1)(3k+2) \\ &\leadsto \text{divisible by } 3 \\ \text{For } n &= 3k+1, \\ n(n+1)(n+2) &= (3k+1)(3k+2)(3k+3) \\ &= \color{red}3 \times (3k+1)(3k+2)(k+1) \\ &\leadsto \text{divisible by } 3 \\ \text{For } n &= 3k+2, \\ n(n+1)(n+2) &= (3k+2)(3k+3)(3k+4) \\ &= \color{red}3 \times (3k+2)(k+1)(3k+4) \\ &\leadsto \text{divisible by } 3 \\ \therefore n(n+1)(n+2) &\text{ is divisible by 3} \end{align} \)
(b) Hence, prove \( n^3 + 2n \) is divisible by \( 3 \) algebraically.
\( \require{AMSsymbols} \begin{align} n^3+2n &= n^3+3n^2+2n-3n^2 \\ &= n\left(n^2+3n+2\right)-3n^2 \\ &= n(n+1)(n+2)-3n^2 \\ &\leadsto \text{divisible by 3} &\color{green}{n(n+1)(n+2) \text{ is divisible by 3 and } 3n^2 \text{ is divisible by 3}} \end{align} \)
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