# Tagged Questions

For questions about mathematical induction, a method of mathematical proof. Mathematical induction generally proceeds by proving a statement for some integer, called the *base case*, and then proving that if it holds for one integer then it holds for the next integer. This tag is primarily meant ...

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### elementary prove thru induction - dumb stumbling

i am trying to prove this statement for all $n \in \mathbb{N}$ with the help of induction: $4 \sum_{k=1}^{n} (-1)^kk=(-1)^n(2n+1)-1$ base case: n=1 $4 \sum_{k=1}^{1} (-1)^11=-4=(-1)^1(2*1+1)-1$ .. ...
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### Induction - Countable Union of Countable Sets

Stephen Abbott has a an exercise in Chapter 1 (1.2.12) that suggests that one cannot use induction to prove that a countable union of countable sets is countably infinite. One answer is that n=...
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### Choosing values in a strong induction

The sequence s0,s1,s2... is defined by s0=1 and for all integers n>0, $s(n)=s(⌊n/2⌋)+s(⌊2n/5⌋) + n.$ Prove, using strong induction, that S(n) > 4n for all integers n>=3. To my knowledge, I only have ...
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### Are there any proofs that only exist by induction?

I've come to learn more about induction recently for proving things, and one thing stands out to me. It seems like you could just data-mine patterns and guess a relationship you think might be ...
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### Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$

My question is: Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer. I'm stuck at the basis step. If I ...
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### Proof of inequality $2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})$ using induction
Prove that $2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})$ if $n \ge 1$ using induction. Can someone help me with this problem please. Base case is easily shown, and ...