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$x_1, x_2, ..., x_{n}$ satisfies $$1^j x_1 + 2^j x_2 + \dots + n^j x_{n} = {(n+1)}^j,$$ where $j = 1, 2, ..., n$.

Find $x_1 + x_2 + \dots + x_{n}$ and $1^{n+1}x_1 + 2^{n+1}x_2 + \dots + n^{n+1}x_{n}$.

My thoughts: the given constraints can be expressed in a matrix


$$A= \begin{bmatrix} \frac{1}{n+1} & \frac{2}{n+1} & ... & \frac{n}{n+1} \\ (\frac{1}{n+1})^2 & (\frac{2}{n+1})^2 & ... & (\frac{n}{n+1})^2 \\ ... \\ (\frac{1}{n+1})^{n} & (\frac{2}{n+1})^{n} & ... & (\frac{n}{n+1})^{n} \\ \end{bmatrix} $$

Then $X = A^{-1} [1, 1,..., 1]^{T}$ but i don't see any special properties with this matrix. Trying the elimination doesn't lead me anywhere either..

Thorough some experiments from smaller $n$, I can find that $x_i=(-1)^{i+1}{n \choose i}$, and by observing the original equations, it does seem to work, but I'd like to find a way to derive the answers, rather than guessing and verifying it.

dxiv
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CuriousMind
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    Why not try something smaller, like with $3$ instead of $2022$, see if anything stands out. – Gerry Myerson Jan 07 '22 at 22:26
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    Problems involving the current year frequently appear in contests. What is the source of this problem? – lulu Jan 07 '22 at 22:26
  • @lulu It's a "daily brainteaser exercise" problem of a social media group I am in. I doubt any contest is being held this time of the year.. – CuriousMind Jan 07 '22 at 22:30
  • The matrix $A$ is almost a Vandermonde matrix. Perhaps it helps you find useful properties? – Jose Avilez Jan 07 '22 at 22:43
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    Try what Gerry suggested above (Simplify the problem, solve and then generalize). Do the equivalent problem for small $n=2,3,4,\ldots$, e.g. for $n=3$ the matrix is $\pmatrix{1/3 & 2/3\1/3^2 & (2/3)^2}$. Find the $x_i$'s and write them down. If you do this for $n=2,3,4$ you will see a pattern (hint: Pascals triangle) that allows you to formulate a hypothesis for what the general solution should be - and thereby the sum - and then you can try to prove that. – Winther Jan 07 '22 at 23:02
  • @Winther thanks, i think I found the answers using your suggestion, but now I am interested in actually "deriving" the answer, not "observing and verify" it – CuriousMind Jan 09 '22 at 00:42
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    Good. If you found the solution (which, IIRC, looks roughly correct though I remember having something like $(-1)^{n+i}$ instead of $(-1)^{i+1}$ but I could remember it wrong) then proving it requires showing that $AX = (1,1,\ldots,1)^T$ so your problem is reduced to proving some sum-identity involving the binomial coefficients ala $\sum_{i=1}^n x_i i^j = (n+1)^j$ for $j=1,2,\ldots,n$. – Winther Jan 09 '22 at 01:08
  • For that: this might be useful https://math.stackexchange.com/questions/3833276/if-px-is-any-polynomial-of-degree-less-than-n-show-that-sum-j-0n-1/3833675#3833675 – Winther Jan 09 '22 at 01:25
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    @Winther Think it's actually $,(-1)^{n+i}, \binom{n+1}{i},$ so you remembered that part right. – dxiv Jan 09 '22 at 07:14

1 Answers1

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but I'd like to find a way to derive the answers, rather than guessing and verifying it

The following is one such possible way, which also avoids calculating the $\,x_j\,$ explicitly.

Let $\,a_j = \sum_{k=1}^nk^j \,x_k\,$ for $\,j \ge 0\,$, with $\,a_j = (n+1)^j\,$ for $\,j=1,2,\dots,n\,$.

The form of $\,a_j\,$ indicates that it satisfies an $\,n^{th}\,$ order linear recurrence with characteristic polynomial $\,p(x)=(x-1)(x-2)\dots(x-n)\,$, which can be expanded to $\,p(x)=\sum_{j=0}^n c_j\,x^j\,$ with $\,c_n=1\,$ and $\,c_0=(-1)^n\,n!\;$ ($\,\dagger\,$). It follows that $\,\sum_{j=0}^n c_j\,a_{p+j} = 0\,$ for all $\,p \ge 0\,$, then:

  • equation for $\,p=0\,$ gives $\,a_0=x_1+x_2+\dots+x_n\,$:

$$ a_0 = - \frac{1}{c_0}\sum_{j=1}^{n} c_j\,a_j = - \frac{1}{c_0}\sum_{j=1}^{n} c_j\,(n+1)^j = -\frac{1}{c_0}\left(p(n+1)-c_0\right)= 1-\frac{n!}{(-1)^n\,n!} \\ = 1-(-1)^n $$

  • equation for $\,p=1\,$ gives $\,a_{n+1}=1^{n+1}x_1 + 2^{n+1}x_2 + \dots + n^{n+1}x_{n}\,$ after similar calculations.

($\,\dagger\,$) $\;$ The polynomial $\,p(x)=(x-1)_n\,$ is a falling factorial and the coefficients $\,c_j=s(n,j)\,$ are Stirling numbers of the first kind, though the answer did not require or use that.

dxiv
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