Monday, June 17, 2019

calculus - prove that for every natural n, $5^n - 2^n$, can be divided by 3


How to prove, using recursion, that for every natural n:$$5^n - 2^n$$ can be divided by 3.


Answer



  1. setting $n=1$, $\implies 5^1-2^1=3$ is divisible by $3$

Thus, the number $5^n-2^n$ is divisible by $3$ for $n=1$



  1. assume for $n=k$, the number $5^n-2^n$ is divisible by $3$ then $$\color{blue}{5^k-2^k=}\color{blue}{3m}$$ where, $m$ is some integer




  2. setting $n=k+1$, $$5^{k+1}-2^{k+1}=5\cdot 5^k-2\cdot 2^k$$ $$=5\cdot 5^k-5\cdot 2^k+3\cdot 2^k$$ $$=5(\color{blue}{5^k-2^k})+3\cdot 2^k$$ $$=5(\color{blue}{3m})+3\cdot 2^k$$ $$=3(5m+2^k)$$ since, $(5m+2^k)$ is an integer hence, the above number $3(5m+2^k)$ is divisible by $3$




Hence, $5^n-2^n$ is divisible by $3$ for all integers $n\ge 1$


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