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

Given that:

$x=\frac{1.3}{3.6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+\frac{\mathrm{1.3.5.7}}{\mathrm{3.6.9.12}}+...........$

To prove:

$9x^2+24x=11$

Solution:

$x=\frac{1.3}{3.6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+\frac{\mathrm{1.3.5.7}}{\mathrm{3.6.9.12}}+...........$

$=\frac{1.3}{2!}{\left(\frac{1}{3}\right)}^2+\frac{\mathrm{1.3.5}}{3!}{\left(\frac{1}{3}\right)}^3+\frac{\mathrm{1.3.5.7}}{4!}{\left(\frac{1}{3}\right)}^4+...........$

Adding ,$1+1\left(\frac{1}{3}\right)$ on both sides we get,

$1+1\left(\frac{1}{3}\right)+x=1+1\left(\frac{1}{3}\right)+\frac{1.(1+2)}{2!}{\left(\frac{1}{3}\right)}^2+\frac{1.(1+2).(1+2.2)}{3!}{\left(\frac{1}{3}\right)}^3+...........$

Here, $p=1,q=2,\frac{x}{q}=\frac{1}{3}$

$⟹\frac{x}{2}=\frac{1}{3}$

$⟹x=\frac{2}{3}$

Since, $(1-x{)}^{\frac{p}{q}}=1+p\left(\frac{x}{q}\right)=\frac{p.(p+q)}{2!}{\left(\frac{x}{q}\right)}^2+\frac{p.(p+q)(p+2q)}{3!}{\left(\frac{x}{q}\right)}^3+.....$

$\therefore \frac{4}{3}+x={(1-\frac{2}{3})}^{\frac{-1}{2}}$

or, $\frac{4+3x}{3}={\left(\frac{1}{3}\right)}^{\frac{-1}{2}}$

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

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

or,$16+24x+9x^2=27$

or, $9x^2+24x=11$