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As the final part of a big proof I got for uni homework: (It is an extra question, may be unsolvable)

$$k^n<\sum_{i=0}^n\binom{n}ik^{n-i}(2^i-1)$$

My idea is to develop the right side into an $(x+y)^n$ type thing, but it is not in a correct form for this.

This is said (by my homeworks) to be true for each $n > 2$, and $k \in N+$

How can I prove/develop this equation?

Edit:

  • I tried induction, but failed to find a common expression.

  • I tried simplifying into $(x+y)^n$ but failed doing that as well, because $(2^i-1)$ is not a $y^i$

  • This started from $k^n < (k+2)^n - (k+1)^n$

Amit
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    if it works for n, does it follow it works for n+1? – shai horowitz May 20 '16 at 20:33
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    There's a subtraction on the right hand side. Use that to write the entire sum as the difference between two sums. Are the two sums now in the "correct form" for what you want? – Dan Piponi May 20 '16 at 20:36
  • @shaihorowitz Tried induction, but failed getting a common expression, because of the combinatoric part (n choose i) – Amit May 20 '16 at 20:36
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    @DanPiponi That is a point before. k^n < (k+2)^n - (k+1)^n, which developed to this, which I don't know how to develop further. – Amit May 20 '16 at 20:37
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    As an aside, are there meant to be limits on what values of $k$ are allowed? I notice that for $k=-1$ and $n$ even, one has $1<1$ which is not true. – JMoravitz May 20 '16 at 20:46
  • @JMoravitz Oh sorry, yes. k is N+ – Amit May 20 '16 at 20:48
  • $$\sum_{i=0}^n\binom nik^{n-1}<\sum_{i=0}^n\binom nik^{n-1}(2i-1)$$ – Simply Beautiful Art May 20 '16 at 20:55
  • @SimpleArt As far as I know, $(x+y)^n$ requires $x > 0, y > 0$, therefore, this cannot be created – Amit May 20 '16 at 20:59
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    For the last line in your question, $12^3-11^3 = 397 \not> 10^3$, is the question correct? – peterwhy May 20 '16 at 21:04
  • This makes me question the supposed proof mentioned by the OP in "it does work for n=3, that was simple to prove." Considering that counterexamples exist for $n=3$, it should not have been possible to prove. I would expect as well that it is false for larger $n$ as well. – JMoravitz May 20 '16 at 21:11
  • @JMoravitz I'm so dumb! looking at the proof I switched the sign (< to >).. Sorry for wasting your time! – Amit May 20 '16 at 21:16
  • @Amit well, even that would be hard, since $5^3 = 125<127 = 7^3-6^3$ – JMoravitz May 20 '16 at 21:17

2 Answers2

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The statement wishing to be proven is as mentioned equivalent to whether or not for $n>2$ and $k\in\Bbb N$ we have the following relation:

$$k^n < (k+2)^n - (k+1)^n$$

This is false.

Consider the validity of the statement for $k=6, n=3$. One has $6^3=216$ and $(6+2)^3 - (6+1)^3 = 512 - 343 = 169$

$$6^3 = 216> 169 = (6+2)^3 - (6+1)^3$$

Why these are equivalent is because $\sum\limits_{i=0}^n \binom{n}{i}k^{n-i}(2^i-1) = \left(\sum\limits_{i=0}^n \binom{n}{i}k^{n-i}2^i\right) - \left(\sum\limits_{i=0}^n\binom{n}{i}k^{n-i}\right) = (k+2)^n - (k+1)^n$

JMoravitz
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On the bright side $$\sum_{i=0}^{n}\binom{n}{i}k^{n-i}\left ( 2^i-1 \right )=\sum_{i=0}^{n}\binom{n}{i}k^{n-i}2^i - \sum_{i=0}^{n}\binom{n}{i}k^{n-i}=(k+2)^n-(k+1)^n=$$$$(k+2 - (k+1))\cdot((k+2)^{n-1}+(k+2)^{n-2}(k+1)+...+(k+1)^{n-1})>$$$$(k+1)^{n-1}+(k+1)^{n-1}+..+(k+1)^{n-1}=n(k+1)^{n-1}$$

This is definitely true for $n>k$.

rtybase
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  • Can you please explain how you did every step? (just formula name, because I'm lost) – Amit May 20 '16 at 21:12
  • @Amit $x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\dots+xy^{n-2}+y^{n-1})$. This is seen to be true via properties of telescoping series. Try it for yourself to see further. There is no "formula name" to my knowledge for this tidbit. See for example, this question and answer. – JMoravitz May 20 '16 at 21:14
  • Yep $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1})$ it's a well known identity. – rtybase May 20 '16 at 21:16
  • @rtybase Thanks. I now understand what you did there – Amit May 20 '16 at 21:21