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We consider all k-element sequences with values from the set $[n] = \{1,2 ,..., n \}$. For each such list, we determine the smallest value, and then sum these values. Show that the sum is equal to:

$$1^k + 2^k + 3^k + ... + n^k$$

Example, for $k = 2$ and $n = \{ 1,2 \}$:

In that case, get $4$ seqences. Those are: $11$, $22$, $12$, $21$. Therefore, we take smallest values from each of those $4$ sequences. These will be: $1,2,1,1$. We sum them up and get $5$.

Why would that be the case?

I know that there are $n^k$ k-element sequences with values from the set $[n]$. From each of those we need to take the smallest value. It will be $1$ for $n^k - (n-1)^{k}$ of those sqeuences (we take all possible sequences and subtract those without any $1$). Therefore we have $(n^k - (n-1)^{k}) \cdot 1$.

  • For $2$ we have: $(n-1)^k - (n-2)^{k}$ sequences because we can take all possible sequences without $1$ and subtract those without any $1$ and $2$
  • And so on...

Therefore we get:

$$\big(n^k - (n-1)^{k} \big) \cdot 1 + \big( (n-1)^k - (n-2)^{k} \big) \cdot 2 + ... + \big( 1^k - 0^{k} \big) \cdot n$$

But that doesn't look like: $1^k + 2^k + 3^k + ... + n^k$, so I don't know what to do...

thefool
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  • For $(n,k)=(2,2)$ there is only one sequence we can take, namely ${1,2}$, right? the least element of that is $1$ so the answer in that case is $1$. But you claim it should be $1^2+2^2=5$. Am I misunderstanding something? More broadly, the case $(n,n)$ always gives $1$ if I have understood the rules correctly. – lulu Sep 08 '23 at 13:37
  • And why do you say there are $n^k$ subsequences ? Surely there are $\binom nk$. – lulu Sep 08 '23 at 13:40
  • No, we take all possible sequences made of $k$ elements with values in set $n$. Therefore, if we have $k = 2$ and $n = {1,2 }$, we get $4$ seqences. Those are: $11$, $22$, $12$, $21$. Then we take smallest values from each of those $4$ sequences. These will be: $1, 2, 1, 1$. We um them up and get $5$, which works fine with the formula stated above. I'm sorry, maybe my description was not clear enough. @lulu – thefool Sep 08 '23 at 15:17

2 Answers2

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As you observed, the number of sequences with smallest term equel to $i$ is $$S_i=(n+1-i)^k-(n-i)^k.$$ It follows that $$\sum_{i=1}^niS_i=\sum_{i=1}^ni(n+1-i)^k-\sum_{i=1}^ni(n-i)^k$$ $$=\sum_{i=0}^{n-1}(i+1)(n-i)^k-\sum_{i=0}^{n-1}i(n-i)^k$$ $$=\sum_{i=0}^{n-1}(n-i)^k=1^k+2^k+\cdots+n^k.$$

user14111
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  • Looks good, but why: $\sum^n_{i=1}i(n-i)^k = \sum^{n-1}_{i=0}i(n-i)^k $ ? @user14111 – thefool Sep 09 '23 at 01:12
  • Because $i(n-i)^k=0$ when $i=0$ and also (since we assume $k\gt0$) when $i=n$, the terms indexed by $i=0$ and $i=n$ can be included or omitted without affecting the sum. – user14111 Sep 09 '23 at 02:01
  • $$\sum_{i=1}^ni(n-i)^k=\sum_{i=1}^{n-1}i(n-i)^k+0=0+\sum_{i=1}^{n-1}i(n-i)^k=\sum_{i=0}^{n-1}i(n-i)^k$$ – user14111 Sep 09 '23 at 02:10
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This looks like a probability problem in disguise. Let $f(n,k)$ denote the sum you are interested in, and let $E(n,k)$ denote the expected value of the minimum of $k$ independent rolls of a fair $n$-sided die with sides numbered from $1$ to $n$. Plainly, $$E(n,k)=\frac{f(n,k)}{n^k}.$$ In my answer to another question I used the method of indicator variables to derive the formula $$E(n,k)=n^{-k}\sum_{i=1}^ni^k$$ from which it follows that $$f(n,k)=n^kE(n,k)=\sum_{i=1}^ni^k.$$ However, your straightforward solution (completed in @user14111's answer) seems simpler.

bof
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