You get your induction started by checking that $a_{n_0}>b_{n_0}$. For the induction step there are two very natural things to try. Compare $a_{k+1}-a_k$ with $b_{k+1}-b_k$: if $a_k>b_k$ and $a_{k+1}-a_k\ge b_{k+1}-b_k$, then clearly $a_{k+1}>b_{k+1}$.

So, if I hand you any $C$ and $n_0$, you have to find $n \geq n_0$ so that $3^n > C \,2^n$. If it isn't clear how to prove it yet, try finding an appropriate $n$ if I give you $C = 700$ and $n_0 = …

自然数立方和公式如下： 1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\frac{1}{4} n^{2}(n+1)^{2} 简记： \sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2 那么这个公式是怎么得到的呢？ 下面来说道说道 【方法一】 根据 k…

2016年12月22日 · The question is prove by induction that $n^3 < 3^n$ for all $n\ge4$. The way I have been presented a solution is to consider: $$\frac{(d+1)^3}{d^3} = (1 + \frac{1}{d})^3 \ge (1.25)^3 = (\frac...

2015年12月8日 · If $n$ is an integer, $3^n > n^3$ unless $n$ = 3. That's easy to prove if n is a negative integer, 0, 1 or 2. For $n$ = 3, $3^n = 3^3 = n^3$. Using that as my base case, I now prove by mathematical induction that $3^n > n^3$ if $n$ is any integer greater than 3. Suppose $3^n \geqslant n^3$ is true for $n = k$, where $k$ is an integer and is at ...

an+1/an=3[n/(n+1)]^n 当n趋于无穷大时，比值=3*e^[-n/(n+1)]=3/e>1， 可知原级数是发散的。

2009年9月15日 · 3^n > n^3 [/tex] You want to show [tex] A_n [/tex] is true for [tex] n \ge 4 [/tex]. It shouldn't be hard to show [tex] A_4 [/tex] is true. Assume it is true for [tex] k \ge 4 [/tex]. Now [tex] 3^{3+1} = 3 \cdot 3 ^k > 3 \cdot k^3 = k^3 + k^3 + k^3 [/tex]

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2018年3月17日 · (n!)/((n-3)!)=n^3-3n^2+2n (n!)/((n-3)!) = (n(n-1)(n-2)color(blue)((n-3)(n-4).....3*2*1))/((n-3)(n- 4).....3*2*1) = n(n-1)(n-2) = n(n^2-3n+2) = n^3-3n^2+2n