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 178400 by infinityaction last updated on 16/Oct/22

$$\mathrm{evaluate}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\mathrm{k}{\mathrm{e}}^{\mathrm{kx}}$$

Answered by mr W last updated on 16/Oct/22

$$S=\underset{k=1}{\sum ^n}{ke}^{kx}=e^x+2e^{2x}+3e^{3x}+...+{ne}^{nx}$$$$e^{x}S=e^{2x}+2e^{3x}+3e^{4x}+...+{ne}^{(n+1)x}$$$$(1-e^x)S=e^x+e^{2x}+e^{3x}+...e^{nx}-{ne}^{(n+1)x}$$$$(1-e^x)S=\frac{e^x(1-e^{nx})}{1-e^x}-{ne}^{(n+1)x}$$$$\Rightarrow S=\frac{e^x(1-e^{nx})}{{(1-e^x)}^2}-\frac{{ne}^{(n+1)x}}{1-e^x}(e^x\ne 1)$$

Commented by infinityaction last updated on 16/Oct/22

$$thankssir$$

Commented by haladu last updated on 16/Oct/22

$$\mathbf{Nice}!\mathbf{solution}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com