Question

If $y=\frac{7}{5}(1+\frac{1}{{10}^2}+\frac{1\cdot 3}{1\cdot 2}\frac{1}{{10}^4}+\frac{1\cdot 3\cdot 5}{1\cdot 2\cdot 3}\frac{1}{{10}^6}+\cdots \infty )$ then $4y^2=$

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 $y=\frac{7}{5}(1+\frac{1}{{10}^2}+\frac{1\cdot 3}{1\cdot 2}\frac{1}{{10}^4}+\frac{1\cdot 3\cdot 5}{1\cdot 2\cdot 3}\frac{1}{{10}^6}+\cdots \infty )$
$=\frac{7}{5}(1+\frac{1}{{10}^2}+\frac{1\cdot (1+2)}{1\cdot 2}\frac{1}{{10}^4}+\frac{1\cdot (1+1\cdot 2)\cdot (1+2\cdot 2)}{1\cdot 2\cdot 3}\frac{1}{{10}^6}+\cdots \infty )$

$=\frac{7}{5}(1+\frac{\frac{1}{2}}{1!}\left(\frac{2}{{10}^2}\right)+\frac{\frac{1}{2}\cdot (\frac{1}{2}+1)}{2!}{\left(\frac{2}{{10}^2}\right)}^2+\frac{\frac{1}{2}\cdot (\frac{1}{2}+1)\cdot (\frac{1}{2}+2)}{3!}{\left(\frac{2}{{10}^2}\right)}^3+\cdots )$
$=\frac{7}{5}\left[{(1-\frac{2}{100})}^{-1/2}\right]$
$=\frac{7\cdot 10}{5\cdot \sqrt{98}}=\sqrt{2}$
$\therefore 4y^2=8$