what i have tried, writing the divident as $(nx^n)(x-1)+(1-x^n)$ then if n is even, we can write the quotient when divided by x-1 as, $nx^n-(x^{n/2}+1)(x^{n/4}+1)....(x+1)$ . i dont know how to divide this again by $x-1$ . if n is odd, we could use $a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+...b^{n-1})$ to write the quotient when divided by $(x-1)$ as $nx^n-(1+x+x^2+...+x^{n-1})$. once again, i dont know how to divide this by $x-1$.
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3Did you try for small $n$ ? Look for some patterns – Claude Leibovici Jun 01 '24 at 07:49
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got it. thanks! – roi Jun 01 '24 at 07:53
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2The answer can be found at https://math.stackexchange.com/questions/2079719/prove-that-the-polynomial-f-nx-nxn1-n1xn1-is-divisible-by-x-12 also https://math.stackexchange.com/questions/1376981/proving-by-induction-sum-limits-k-1nkqk-1-frac1-n1qn-nqn – Gerry Myerson Jun 01 '24 at 07:58
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1It's also buried in https://math.stackexchange.com/questions/544434/monotonicity-in-x-of-fracx-alpha-1x-1 – Gerry Myerson Jun 01 '24 at 08:03
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Just like you mentioned, we can write the dividend polynomial as: $$ p(x)=nx^n(x-1)-(x^n-1) = (x-1)(nx^n-x^{n-1}-\dots-x-1) $$ Let $q(x) = nx^n-x^{n-1}-\dots-x-1$. The quotient can be found by dividing $q(x)$ by $x-1$. We have: $$ q(x) = \sum_{i=1}^{n} x^n-x^{n-i} = \sum_{i=1}^{n} x^{n-i}(x^i-1) $$ So: $$ q(x)/(x-1) = \sum_{i=1}^{n} x^{n-i}(\sum_{j=0}^{i-1}x^j)=\sum_{j<i}x^{n-i+j}=\sum_{k=1}^n (n-k+1)x^{n-k}=\sum_{k=0}^{n-1} (k+1)x^{k} $$
Aref Sadeghi
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@RedFive This is not a new question. The second half is the correct answer (the first half was an answer to another question, by misreading). – Anne Bauval Jun 01 '24 at 09:52
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@Aref your answer would be clearer if you removed the first half, which is off topic. (Up to "Looks like I misread the problem", included.) – Anne Bauval Jun 01 '24 at 09:53
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1@AnneBauval you are correct and I do not know why this message has come up as I thought I clicked on "skip" when reviewing this. – Red Five Jun 01 '24 at 10:14
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For each $n\in\mathbb{N}$, you have \begin{align} (x-1)\left(nx^n-\sum_{k=0}^{n-1}x^k\right)&=nx^{n+1}-\sum_{k=1}^nx^k-nx^n+\sum_{k=0}^{n-1}x^k\\ &=nx^{n+1}-(n+1)x^n+1. \end{align} On the other hand, \begin{align} nx^n-\sum_{k=0}^{n-1}x^k&=\sum_{k=0}^{n-1}(x^n-x^k)\\ &=\sum_{k=0}^{n-1}x^k(x^{n-k}-1)\\ &=(x-1)\sum_{k=0}^{n-1}x^k(1+x+x^2+\cdots~x^{n-k-1})\\ &=(x-1)\sum_{k=0}^{n-1}(x^k+x^{k+1}+\cdots+x^{n-1}). \end{align} Therefore, \begin{align} nx^{n+1}-(n+1)x^k+1&=(x-1)^2\sum_{k=0}^{n-1}(x^k+x^{k+1}+\cdots+x^{n-1})\\ &=(x-1)^2(1+2x+3x^2+\cdots+nx^{n-1}). \end{align}
Another User
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I think the hope was to find an answer in the form $a_0x^{n-1}+a_1x^{n-2}+\dotsb+a_{n-1}$. That's hard to pick out from what you have found. – Gerry Myerson Jun 01 '24 at 08:50
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