Question

If $y=1-\frac{1}{5}+\frac{1\cdot 4}{5\cdot 10}-\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}+\cdots \infty$ then $8y^3=$

Answer 1

If $n\in Q-W$, we know that
$(1-x{)}^{-n}=1+nx+\frac{n(n+1)}{2!}{x}^2+\frac{n(n+1)(n+2)}{3!}{x}^3+\cdots$
Given
$y=1-\frac{1}{5}+\frac{1\cdot 4}{5\cdot 10}-\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}+\cdots$
$=1+\frac{1}{1}\cdot (-\frac{1}{5})+\frac{1\cdot (1+3)}{1\cdot 2}\cdot {(-\frac{1}{5})}^2+\frac{1\cdot (1+3)\cdot (1+2\cdot 3)}{1\cdot 2\cdot 3}\cdot {(-\frac{1}{5})}^3+\cdots$
$=1+\frac{1/3}{1!}(-\frac{3}{5})+\frac{1/3\cdot (1/3+1)}{2!}{(-\frac{3}{5})}^2+\cdots$
$y={(1+\frac{3}{5})}^{-1/3}={\left(\frac{8}{5}\right)}^{-1/3}=\frac{5^{1/3}}{2}$
$\therefore 8y^3=5$