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$$q \ne 0,1,M \in {\Bbb N}{\kern 1pt} $$ Alread konw: $$\sum\limits_{n = 0}^{N - 1} {{q^n}n = \frac{{{q^N}((q - 1)N - q) + q}}{{{{(q - 1)}^2}}}} {\kern 1pt} $$ $$\sum\limits_{n = 0}^{N - 1} {{2^n}n = {2^N}(N - 2) + 2} {\kern 1pt} $$ $$\sum\limits_{n = 0}^{N - 1} {{2^n}{n^2} = {2^N}({N^2} - 4N + 6) - 6} {\kern 1pt} $$ $$\sum\limits_{n = 0}^{N - 1} {{3^n}n = \frac{{{3^N}(2N - 3) + 3}}{4}} {\kern 1pt} $$ $$\sum\limits_{n = 0}^{N - 1} {{3^n}{n^2} = \frac{{{3^N}(4{N^2} - 12N + 12) - 12}}{8}} {\kern 1pt} $$

Is there a formula for general q,M.

RobPratt
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mcfroo kknd
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1 Answers1

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$$\sum\limits_{n = 1}^{N - 1} {{q^n}} \left( {_M^{n + K}} \right) = {q^N}\sum\limits_{g = 0}^M {{{( - 1)}^{^g}}} \frac{{\left( {_{M - g}^{N + K - 1 - g}} \right)}}{{{{(q - 1)}^{g + 1}}}} + \frac{{{q^{M - k}}}}{{{{(1 - q)}^{M + 1}}}}$$ Expand nM $${n^M} = \sum\limits_{g = 0}^M {g!\left\{ {_g^M} \right\}} \left( {_g^n} \right) = \sum\limits_{g = 0}^M {{{( - 1)}^{M - g}}g!\left\{ {_g^M} \right\}} \left( {_g^{n + g - 1}} \right) = \sum\limits_{g = 0}^M {\left\langle {_g^M} \right\rangle } \left( {_M^{n + g}} \right)$$ By combining the two, the formulas can be obtained: $$A_q^M = \sum\limits_{k = 0}^M {{{(1 - q)}^{M - k}}} {q^k}\left\{ {_k^M} \right\}k! = q\sum\limits_{k = 0}^M {{{(q - 1)}^{M - k}}} \left\{ {_k^M} \right\}k! = \sum\limits_{k = 0}^M {\left\langle {_k^M} \right\rangle } {q^{1 + k}}$$

$$\sum\limits_{k = 0}^M {{q^n}} {n^M} = \frac{{{q^N}}}{{{{(q - 1)}^{M + 1}}}}\sum\limits_{g = 0}^M {{{(q - 1)}^g}} {( - 1)^{M - g}}\left( {_g^M} \right)A_q^{M - g}{N^g} + \frac{{A_q^M}}{{{{(1 - q)}^{M + 1}}}}$$

$$|q|<1,\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 0}^{N - 1} {{q^n}} {n^M} = \frac{{A_q^M}}{{{{(1 - q)}^{M + 1}}}} \to A_t^M = tA(M,t)$$ A(M,t) is Eulerian polynomials.See Further Study of the Shape of the Numbers and More Calculation Formulas

More progress at https://www.preprints.org/manuscript/202305.0311/v1

mocooJ
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