Question

$\frac{4}{7}-\frac{5}{7^2}+\frac{4}{7^3}-\frac{5}{7^4}+\frac{4}{7^5}-\frac{5}{7^6}+....$ to infinity.

Answer 1

Following series can be arranged as

$(\frac{4}{7}\times \frac{7}{7}-\frac{5}{7^2})+(\frac{4}{7^3}\times \frac{7}{7}-\frac{5}{7^4})+............\infty$

$=\frac{23}{7^2}+\frac{23}{7^4}+\frac{23}{7^6}............\infty$

Common ratio = $r=\frac{1}{7^2}<1$

If $n=\infty$ & $\left|r\right|<1$ then formula for total sum is

$S=\frac{a}{1-r}$

$\Rightarrow S=\frac{\frac{23}{7^2}}{1-\frac{1}{7^2}}=\frac{23}{48}$