We prove that for all $n \in \mathbb{N}$, $4 \mid \left( 5^n-1 \right)$. (Notationally, this says $4$ divides $5^n-1$ with a zero remainder). For a basis, let $n=1$.

Define f(n) to be as a function on the natural numbers by: f(n) = n/2 if n is even and f(n) = 3n + 1 if n is odd. The conjecture is that for all n ∈N, n eventually converges under iteration by f to 1. I was wondering if the "5n+1" problem has been solved.

2021年3月7日 · 验证一下，对于 n^5 当 n=0 时， n^{5}-n=0 是 5 的倍数. 那么对于 \left( n+1 \right)^5 有 \left( n+1 \right)^{5}=n^{5}+C_{5}^{1}n^{4}+...+C_{5}^{4}n+1. 为了凑出 \left( n+1 \right)^{5}-\left( n+1 \right) ，等式两边同时减去 \left( n+1 \right) ，得到

Base case: $n=1$ It's easy to see that $5^n -4n-1$ is divisible by $16$ because for $n=1$: $$5^n -4n-1 = 5 - 4 - 1 = 0$$ and $0$ is divisible by $16$. Induction hypothesis : assume now that $5^n -4n-1$ is divisible by $16$ for an arbitrary $n$

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\sum_{n=0}^{\infty}\frac{3}{2^n} Show More; Description. Solve problems from Pre Algebra to Calculus step-by-step step-by-step. expand 5^{n-1} en. Related Symbolab blog posts. Practice Makes Perfect. Learning math takes practice, lots of practice. Just like running, it takes practice and dedication. If you want...