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Question Number 37930 by zesearho last updated on 19/Jun/18

$$n\in \mathbb{N}$$$$U_{n+1}={\left(\frac{1}{2}\right)}^{n+1}+U_n$$$$U_n=?$$

Commented by abdo.msup.com last updated on 19/Jun/18

$$wehave{u}_{n+1}-u_n={\left(\frac{1}{2}\right)}^{n+1}\Rightarrow$$$$\sum _{k=0}^{n-1}(u_{k+1}-u_k)=\sum _{k=0}^{n-1}{\left(\frac{1}{2}\right)}^{k+1}\Rightarrow$$$$u_n-u_0=\sum _{k=1}^n{\left(\frac{1}{2}\right)}^k=\frac{1-{\left(\frac{1}{2}\right)}^{n+1}}{1-\frac{1}{2}}-1$$$$=2-2{\left(\frac{1}{2}\right)}^{n+1}-1=1-\frac{1}{2^n}\Rightarrow$$$$u_n=u_0+1-\frac{1}{2^n}.$$

Answered by ajfour last updated on 19/Jun/18

$$U_1=U_0+\frac{1}{2}$$$$U_2=U_1+{\left(\frac{1}{2}\right)}^2$$$$U_n=U_0+1-{\left(\frac{1}{2}\right)}^n.$$

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