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Question Number 69566 by Ajao yinka last updated on 25/Sep/19

Answered by mind is power last updated on 25/Sep/19

$$=\underset{x\to +\mathrm{\infty }}{\lim }\sum _{k=0}^x{\int }_k^{k+1}{(x-k)}^k{e}^{-nk}dx$$$$=\underset{x\to +\mathrm{\infty }}{\lim }\sum _{k=0}^x{e}^{-nk}{\int }_k^{k+1}{(x-k)}^{k}dx$$$$=\underset{x\to +\mathrm{\infty }}{\lim }\sum _{k=0}^x{e}^{-nk}{\left[\frac{{(x-k)}^{k+1}}{k+1}\right]}_k^{k+1}$$$$=\underset{x\to +\mathrm{\infty }}{\lim }\sum _{k=0}^x{e}^{-nk}.\frac{1}{k+1}$$$$A=\underset{x\to +\mathrm{\infty }}{\lim }\sum _{k=0}^x{e}^{-(k+1)n}.\frac{e^n}{k+1}$$$$$$$$wehave-\ln (1-x)=\underset{k=0}{\sum ^{+\mathrm{\infty }}}\frac{x^{k+1}}{k+1}$$$$A=e^n\underset{k=0}{\sum ^{+\mathrm{\infty }}}\frac{{\left(e^{-n}\right)}^{k+1}}{k+1}=e^n\times -ln(1-e^{-n})=e^{n}ln\left(\frac{1}{1-e^{-n}}\right)=e^n\ln \left(\frac{e^n}{e^n-1}\right)=e^n(n-ln(e^n-1))$$$$$$

Commented by Ajao yinka last updated on 26/Sep/19

$$perfectsolution$$

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