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# need help

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If $$\displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{n} + \frac{1}{62} + \frac{1}{126} + \frac{1}{248} = \frac{495}{496}$$ what is the value of n?

May 15, 2020

#1
+24909
+1

If

$$\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{n} + \dfrac{1}{62} + \dfrac{1}{126} + \dfrac{1}{248} = \dfrac{495}{496}$$

what is the value of $$n$$?

$$\begin{array}{|rcll|} \hline \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{n} + \frac{1}{62} + \frac{1}{126} + \frac{1}{248} &=& \frac{495}{496} \quad | \quad \times 2 \\ \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \frac{2}{n} + \frac{2}{62} + \frac{2}{126} + \frac{2}{248} &=& \frac{495}{248} \\ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{2}{n} + \frac{1}{31} + \frac{1}{63} + \frac{2}{248} &=& \frac{495}{248} \\ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{2}{n} + \frac{1}{31} + \frac{1}{63} &=& \frac{495}{248} - \frac{2}{248} \\ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{2}{n} + \frac{1}{31} + \frac{1}{63} &=& \frac{493}{248} \quad | \quad \times 8 \\ 8 + \frac{8}{2} + \frac{8}{4} + \frac{8}{8} + \frac{16}{n} + \frac{8}{31} + \frac{8}{63} &=& \frac{493}{31} \\ 8 + 4 + 2 + 1 + \frac{16}{n} + \frac{8}{31} + \frac{8}{63} &=& \frac{493}{31} \\ 15 + \frac{16}{n} + \frac{8}{63} &=& \frac{493}{31} - \frac{8}{31} \\ 15 + \frac{16}{n} + \frac{8}{63} &=& \frac{485}{31} \\ \frac{16}{n} + \frac{8}{63} &=& \frac{485}{31} -15 \\ \frac{16}{n} + \frac{8}{63} &=& \frac{485-15*31}{31} \\ \frac{16}{n} + \frac{8}{63} &=& \frac{20}{31} \\ \frac{16}{n} &=& \frac{20}{31} - \frac{8}{63} \\ \frac{16}{n} &=& \frac{20*63-8*31}{31*63} \\ \frac{16}{n} &=& \frac{1012}{1953} \\ \frac{1}{n} &=& \frac{1012}{16*1953} \\ \frac{1}{n} &=& \frac{1012}{31248} \\ \frac{1}{n} &=& \frac{253}{7812} \\ \mathbf{n} &=& \mathbf{\frac{7812}{253}} \\ \hline \end{array}$$

May 15, 2020