2^n(n^2)

相關問題 & 資訊整理

2^n(n^2)

Prove that for all n ≥ 5, 2n > n2. BASIS: When n = 5, 2n = 25 = 32 and n2 = 52 = 25. As 32 > 25 the statement is true for ... ,It can be replaced (in both places it occurs) by any other integer, and the Principle still works. Example 1. Prove that for all n ≥ 5, 2n > n2. BASIS: When n = 5, 2n = ... ,You proved it's true for n=5. Now suppose it's true for some integer n≥5. The aim is to prove it's true for n+1. But (♤)2n+1=2×2n>2×n2>(n+1)2. ,First since one must have n≠3, the induction base must be n=4. For the induction step: Suppose n2≤2n. Then, (n+1)2=n2+2n+1≤2n+2n+1≤2n+2n=2n+1. ,n2<2n⟺2logn<nlog2⟺lognn<log22. But we know that lognn→n→∞0 , so the above inequality's definitely true from one definite index n and on...but not for all ... ,Prove that $$2^n > n^2$$ if n is an integer greater than 4. ... 4k>2k+1 holds in general for all real numbers k>1/2 (if you're unsure about this, then draw the ...

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2^n(n^2) 相關參考資料
Mathematical Induction Principle of Mathematical Induction ...

Prove that for all n ≥ 5, 2n &gt; n2. BASIS: When n = 5, 2n = 25 = 32 and n2 = 52 = 25. As 32 &gt; 25 the statement is true for ...

http://www.math.uvic.ca

Notes on Mathematical Induction

It can be replaced (in both places it occurs) by any other integer, and the Principle still works. Example 1. Prove that for all n ≥ 5, 2n &gt; n2. BASIS: When n = 5, 2n = ...

https://www.math.uvic.ca

Proof by induction: $2^n &gt; n^2$ for all integer $n$ greater than ...

You proved it's true for n=5. Now suppose it's true for some integer n≥5. The aim is to prove it's true for n+1. But (♤)2n+1=2×2n&gt;2×n2&gt;(n+1)2.

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Proof of $n^2 leq 2^n$. - Mathematics Stack Exchange

First since one must have n≠3, the induction base must be n=4. For the induction step: Suppose n2≤2n. Then, (n+1)2=n2+2n+1≤2n+2n+1≤2n+2n=2n+1.

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Proof that $n^2 &lt; 2^n$ - Mathematics Stack Exchange

n2&lt;2n⟺2logn&lt;nlog2⟺lognn&lt;log22. But we know that lognn→n→∞0 , so the above inequality's definitely true from one definite index n and on...but not for all ...

https://math.stackexchange.com

Prove that $$2^n &gt; n^2$$ if n is an integer greater than 4 ...

Prove that $$2^n &gt; n^2$$ if n is an integer greater than 4. ... 4k&gt;2k+1 holds in general for all real numbers k&gt;1/2 (if you're unsure about this, then draw the ...

https://www.slader.com