Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 154495 by mathdanisur last updated on 18/Sep/21

$$\mathrm{if}{\mathrm{S}}_{\mathbf{n}}\left(\mathrm{t}\right)={\mathrm{n}}^{1-\mathbf{t}}(\frac{{(\mathrm{n}+1)}^{2\mathbf{t}}}{{\left(\sqrt[\mathbf{n}+1]{(\mathrm{n}+1)!}\right)}^{\mathbf{t}}}-\frac{{\mathrm{n}}^{2\mathbf{t}}}{{\left(\sqrt[\mathbf{n}]{\mathrm{n}!}\right)}^{\mathbf{t}}})$$ $$\mathrm{with}\mathrm{t}>0$$ $$\text{Double subscripts: use braces to clarify}$$

Answered by Ar Brandon last updated on 18/Sep/21

$${\mathrm{S}}_{\mathbf{n}}\left(\mathrm{t}\right)={\mathrm{n}}^{1-\mathbf{t}}(\frac{{(\mathrm{n}+1)}^{2\mathbf{t}}}{{\left(\sqrt[\mathbf{n}+1]{(\mathrm{n}+1)!}\right)}^{\mathbf{t}}}-\frac{{\mathrm{n}}^{2\mathbf{t}}}{{\left(\sqrt[\mathbf{n}]{\mathrm{n}!}\right)}^{\mathbf{t}}})$$ $$S_n\left(t\right)=n^{1-t}(\frac{{(n+1)}^{2t}}{{\left(\sqrt[n+1]{\sqrt{2\pi (n+1){\left(\frac{n+1}{e}\right)}^{n+1}}}\right)}^t}-\frac{n^{2t}}{{\left(\sqrt[n]{\sqrt{2\pi n{\left(\frac{n}{e}\right)}^n}}\right)}^t})$$ $$=\frac{1}{n^{t-1}}(\frac{{(n+1)}^{2t}}{{\left(2\pi e^{-1}{(n+1)}^{n+2}\right)}^{\frac{t}{2(n+1)}}}-\frac{n^{2t}}{{\left(2\pi e^{-1}{n}^{n+1}\right)}^{\frac{t}{2n}}})$$ $$=\frac{1}{n^{t-1}}(\frac{{(n+1)}^{2t-\frac{t(n+2)}{2(n+1)}}}{{\left(2\pi e^{-1}\right)}^{\frac{t}{2(n+1)}}}-\frac{n^{2t-\frac{t(n+1)}{2n}}}{{\left(2\pi e^{-1}\right)}^{\frac{t}{2n}}})$$ $$=\frac{1}{n^{t-1}}(\frac{{(n+1)}^{\frac{3tn+2t}{2(n+1)}}}{{\left(2\pi e^{-1}\right)}^{\frac{t}{2(n+1)}}}-\frac{n^{\frac{3tn-t}{2n}}}{{\left(2\pi e^{-1}\right)}^{\frac{t}{2n}}})$$

Commented bymathdanisur last updated on 19/Sep/21

$$\mathrm{Thank}\mathrm{You}\mathbf{S}\mathrm{er},\mathrm{answer}=1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com