Question

121+729
$\frac{3}{2} x ^{2}$

Answer 1

the end of line 4. It should be \[ \begin{split} A_R&=\frac{4}{9}+\frac{20}{81}+\frac{100}{729}+\frac{500}{6561}+\dots \\ &=\frac{4}{3^2}+5\times\frac{4}{3^4}+5\times5\times\frac{4}{3^6}+5\times5\times5\times\frac{4}{3^8}+\dots \\ &=\sum_{n=1}^{\infty}5^{n-1}\times\frac{4}{3^{2n}} \\ &=\sum_{n=0}^{\infty}5^{n-1}\times\frac{4}{3^2n+2} \\ &=\frac{\frac{4}{9}}{1-\frac{5}{9}} \\ &=1 \end{split} \]