Series - Problem 2

$\dfrac{4}{7} - \dfrac{5}{7^2} + \dfrac{4}{7^3} - \dfrac{5}{7^4} + \dfrac{4}{7^5} - \dfrac{5}{7^6}+ ...........\infty$

$=(\dfrac{4}{7} + \dfrac{4}{7^3} + \dfrac{4}{7^5}+.....\infty) - (\dfrac{5}{7^2} + \dfrac{5}{7^4} + \dfrac{5}{7^6}+.....\infty) \\= \dfrac{\dfrac{4}{7}}{1-\dfrac{1}{7^2}} - \dfrac{\dfrac{5}{7^2}}{1-\dfrac{1}{7^2}}\\=\dfrac{7}{12} - \dfrac{5}{48}\\= \dfrac{23}{48}$

$S = \frac {4}{7} - \frac {5}{7^2} + \frac {4}{7^3} - \frac {5}{7^4} + \frac {4}{7^5} - \frac {5^6}{7^2} +...$

$\quad = \frac {28-5}{7^2} + \frac {28-5}{7^4} + \frac {28-5}{7^6} + ...$

$\quad = \frac {23}{49} (1 + \frac {1}{7^2} + \frac {1}{7^4} + \frac {1}{7^6} +... )$

$\quad = \frac {23}{49} ( \frac {1}{1- \frac {1}{49} } ) = \frac {23}{49} ( \frac {1}{\frac {48}{49} } ) = \frac {23}{49} ( \frac {49}{48 } ) = \boxed{\frac {23}{48} }$

Let $S=\frac47-\frac5{7^2}+\cdots$ , then $S=\frac47-\frac5{7^2}+\frac S{7^2}$ therefore making $S$ the subject we get $S=\frac{23}{48}$

Let,s separate the question into 2 parts: $\color{#3D99F6}{\frac{4}{7}+\frac{4}{7^3}+\frac{4}{7^5}+\dotsm)-(\frac{5}{7^2}+\frac{5}{7^4}+\frac{5}{7^6}+\dotsm)}$ Now let,s find sum of each part separately. $\color{#D61F06}{Let\;S=\frac{4}{7}+\frac{4}{7^3}+\frac{4}{7^5}+\dotsm}$ $\color{#69047E}{Then\;\frac{S}{7^2}=\frac{4}{7^3}+\frac{4}{7^5}+\dotsm}$ Subtracting the first equation from the second equation gives us: $\color{#CEBB00}{S-\frac{S}{7^2}=\left(\frac{4}{7}+\frac{4}{7^3}+\dotsm\right)-\left(\frac{4}{7^3}+\frac{4}{7^5}+\dotsm\right)}$ $\color{#EC7300}{\frac{48S}{7^2}=\frac{4}{7}}$ $\color{#20A900}{48S=\frac{4}{7}\times7^2=4\times7=28\rightarrow S=\frac{28}{48}=\frac{7}{12}}$ Now let,s deal with the second part: $\color{#E81990}{Let\;M=\frac{5}{7^2}+\frac{5}{7^4}+\dotsm}$ $\color{#3D99F6}{Then\;\frac{M}{7^2}=\frac{5}{7^4}+\frac{5}{7^6}+\dotsm}$ Subtracting the first equation from the second equation gives us: $\color{#CEBB00}{M-\frac{M}{7^2}=\left(\frac{5}{7^2}+\frac{5}{7^4}+\dotsm\right)-\left(\frac{5}{7^4}+\dotsm\right)}$ $\color{#EC7300}{\frac{48M}{7^2}=\frac{5}{7^2}\rightarrow48m=5\rightarrow M=\frac{5}{48}}$ Now we have the necessary data.Subtracting the second fraction from the first gives the answer: $\frac{28}{48}-\frac{5}{48}=\frac{28-5}{48}=\boxed{\frac{23}{48}}$ I know my solution is too long but this is the way I solved it becuse I wasnt introduced to the formula for sum of geometric series.