Question

1+1/3+1/9+1/27+1/81+1/243+1/729

Answer 1

\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}\)+\(\dfrac{1}{243}+\dfrac{1}{729}\)=\(\dfrac{1093}{729}\)

Answer 2

\(=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

\(\dfrac{729}{729}+\dfrac{243}{729}+\dfrac{81}{729}+\dfrac{27}{729}+\dfrac{9}{729}+\dfrac{3}{729}+\dfrac{1}{729}\)

\(=\dfrac{\left(729+243+81+27+9+3+1\right)}{729}=\dfrac{1084}{729}\)