n-N-U-n-1-1-2-n-1-U-n-U-n-

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

n \in N

U_{n + 1} = {(\frac{1}{2})}^{n + 1} + U_{n}

U_{n} = ?

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

w e h a v e u_{n + 1} - u_{n} = {(\frac{1}{2})}^{n + 1} \Rightarrow

\sum_{k = 0}^{n - 1} (u_{k + 1} - u_{k}) = \sum_{k = 0}^{n - 1} {(\frac{1}{2})}^{k + 1} \Rightarrow

u_{n} - u_{0} = \sum_{k = 1}^{n} {(\frac{1}{2})}^{k} = \frac{1 - {(\frac{1}{2})}^{n + 1}}{1 - \frac{1}{2}} - 1

= 2 - 2 {(\frac{1}{2})}^{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} + {(\frac{1}{2})}^{2}

U_{n} = U_{0} + 1 - {(\frac{1}{2})}^{n} .

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