Question

If $x=\frac{1.3}{3.6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+\frac{\mathrm{1.3.5.7}}{\mathrm{3.6.9.12}}+......,$ then prove that $9x^2+24x=11$.

Answer 1

$\frac{1}{3}.\frac{3}{6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+\frac{\mathrm{1.3.5.7}}{\mathrm{3.6.9.12}}.....$

$={\left(\frac{1}{3}\right)}^2\times \frac{1\times 3}{2!}+{\left(\frac{1}{3}\right)}^3\times \frac{\mathrm{1.3.5}}{3!}+{\left(\frac{1}{3}\right)}^4\times \frac{\mathrm{1.3.5.7}}{4!}.....$

$={\left(\frac{2}{3}\right)}^2\times \frac{1}{2!}\times \frac{1}{2}\times \frac{3}{2}+{\left(\frac{2}{3}\right)}^3\times \frac{\frac{1}{2}\times \frac{3}{2}\times \frac{5}{2}}{3!}+{\left(\frac{2}{3}\right)}^4\times \frac{\frac{1}{2}\times \frac{3}{2}\times \frac{5}{2}\times \frac{7}{2}}{4!}.....$

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

$={(1-\frac{2}{3})}^{\frac{-1}{2}}-(1+\frac{1}{3})$

$=\sqrt{3}-\frac{4}{3}=x$

So $x+\frac{4}{3}=\sqrt{3}$

$3x+4=3\sqrt{3}$

${(3x+4)}^2=27$

$9x^2+24x=11$