# Tagged Questions

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### Better proof that $n \leq 2n$ for all natural numbers?

I tried proving via induction on naturals that $n \leq 2n$ for each natural $n$. Obviously, $0 \leq 2(0)$, and then assuming for any given $n$, $n \leq 2n$, you just show that $n + 1 \leq 2(n + 1).$ ...
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### Is it possible to use mathematical induction to prove a statement concerning all real numbers, not necessarily just the integers? [duplicate]

I am referring to the part of proof by mathematical induction where you show that "if it is true for one value k then it is true for the value k+1". Does proof by induction work over all real numbers? ...
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### Inductive step assumption for all numbers up to $n$

I know that the inductive step should be "for all $n$ (if $P(n)$ then $P(n+1)$)" and NOT "if (for all $n$ $(P(n)$)) then (for all $n$ ($P(n+1)$))" - see this answer. But can it be like "if (for all ...
(b) Prove that for every integer $n \ge 1$, $$1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$$ This is the second part of a two part question. Part (a) was the following: ...