Question

If $x=\frac{5}{(2!)3}+\frac{5\cdot 7}{(3!)3^2}+\frac{5\cdot 7\cdot 9}{(4!)3^3}+.....$, then find the value of $x^2+4x$.

Answer 1

$x=\frac{5}{2!\times 3}+\frac{5.7}{3!\times 3^2}+\frac{\mathrm{5.7.9}}{4!\times 3^3}+...$

$=\frac{3.5}{2!\times 3^2}+\frac{\mathrm{3.5.7}}{3!\times 3^3}+\frac{\mathrm{3.5.7.9}}{4!\times 3^4}+...$

$=\frac{\frac{3}{5}.\frac{5}{2}}{2!}.\frac{2^2}{3^2}+\frac{\frac{3}{5}.\frac{5}{2}.\frac{7}{2}}{3!}.\frac{2^3}{3^3}+\frac{\frac{3}{5}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!}.\frac{2^4}{3^4}+...$

Adding and Subtracting $1+\frac{\frac{3}{2}.\frac{2}{3}}{1!}$

$=[1+\frac{\frac{3}{5}.\left(\frac{2}{3}\right)}{1!}+\frac{\frac{3}{5}.\frac{5}{2}.{\left(\frac{2}{3}\right)}^2}{2!}+\frac{\frac{3}{5}.\frac{5}{2}.\frac{7}{2}.{\left(\frac{2}{3}\right)}^3}{3!}+...]-[1+\frac{\frac{3}{2}.\frac{2}{3}}{1!}]$

$={(1-\frac{2}{3})}^{-\frac{3}{2}}-[1+\frac{\frac{3}{2}.\frac{2}{3}}{1!}]$

$(\because {(1-x)}^{-n}=(1+nx+\frac{n(n+1)}{2!}{x}^2+...))$

$={\left(\frac{1}{3}\right)}^{-\frac{3}{2}}-2=3^{\frac{3}{2}}-2=x$

$\Rightarrow {\left(3\right)}^{\frac{3}{2}}=x+2$

$\Rightarrow {\left({\left(3\right)}^{\frac{3}{2}}\right)}^2={(x+2)}^2$

$27=x^2+4x+4$

$x^2+4x=23$