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Question Number 44527 by Necxx last updated on 30/Sep/18

Commented by Necxx last updated on 30/Sep/18

$$21please$$

Commented by maxmathsup by imad last updated on 30/Sep/18

$$21)let{S}_n=\sum _{k=0}^n{(-1)}^k{C}_n^k\frac{1+k\lambda }{{(1+n\lambda )}^k}with\lambda =ln\left(10\right)$$$$S_n=\sum _{k=0}^n{C}_n^k\frac{{(-1)}^k}{{(1+n\lambda )}^k}+\lambda \sum _{k=0}^n{C}_n^k{(-1)}^k\frac{k}{{(1+n\lambda )}^k}but$$$$\sum _{k=0}^n{C}_n^k{\left(\frac{-1}{1+n\upsilon }\right)}^k={(1-\frac{1}{1+n\lambda })}^n={\left(\frac{n\lambda }{1+n\lambda }\right)}^n$$$$\sum _{k=0}^n{C}_n^{k}k{\left(\frac{-1}{1+n\lambda }\right)}^k?letfind\sum _{k=0}^n{C}_n^k{kx}^k$$$$wehave\sum _{k=0}^n{x}^k=\frac{x^{n+1}-1}{x-1}\Rightarrow \sum _{k=1}^{n}k{x}^{k-1}=\frac{{nx}^{n+1}-(n+1)x^n+1}{{(1-x)}^2}\Rightarrow$$$$\sum _{k=1}^{n}k{x}^k=\frac{x}{{(1-x)}^2}\{{nx}^{n+1}-(n+1)x^n+1\}\Rightarrow$$$$\sum _{k=0}^n{C}_n^{k}k{\left(\frac{-1}{1+n\lambda }\right)}^k=\frac{-1}{(1+n\lambda ){(1+\frac{1}{1+n\lambda })}^2}\{n{\left(\frac{-1}{1+n\lambda }\right)}^{n+1}-(n+1){\left(\frac{-1}{1+n\lambda }\right)}^n+1\}$$$$resttofinichthecalculusthisistheway...$$$$$$

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