Monday, December 17, 2018

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 $n\ge 1$


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...