Question

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

Answer 1

Hint: Rewrite the equation in such a manner that we can figure out the pattern and use the binomial expansion of the expression ${{\left( 1-y \right)}^{-n}}$ to reach the required result.

Complete step-by-step solution -
To start with the solution, we will add 1 and $\dfrac{1}{3}$ to both sides of the equation given in the question. On doing so, we get
$x=\dfrac{1.3}{3.6}+\dfrac{1.3.5}{3.6.9}+\dfrac{1.3.5.7}{3.6.9.12}............$
$x+1+\dfrac{1}{3}=1+\dfrac{1}{3}+\dfrac{1.3}{3.6}+\dfrac{1.3.5}{3.6.9}+\dfrac{1.3.5.7}{3.6.9.12}............$
Now we will rewrite the terms using the property that multiplication is commutative and associative in nature. On doing so, we get
$x+1+\dfrac{1}{3}=1+\dfrac{1}{3}+\dfrac{1.3}{1.2}{{\left( \dfrac{1}{3} \right)}^{2}}+\dfrac{1.3.5}{1.2.3}{{\left( \dfrac{1}{3} \right)}^{3}}+\dfrac{1.3.5.7}{1.2.3.4}{{\left( \dfrac{1}{3} \right)}^{4}}............$
$\Rightarrow x+1+\dfrac{1}{3}=1+\dfrac{1}{3}+\dfrac{1.3}{2!}{{\left( \dfrac{1}{3} \right)}^{2}}+\dfrac{1.3.5}{3!}{{\left( \dfrac{1}{3} \right)}^{3}}+\dfrac{1.3.5.7}{4!}{{\left( \dfrac{1}{3} \right)}^{4}}............$
Here we have taken the common 3 from denominators and have arranged as we have the binomial expansion.
Now we know that the binomial expansion of ${{\left( 1-y \right)}^{-n}}$ expression is:
${{\left( 1-y \right)}^{-n}}=1+ny+\dfrac{n\left( n+1 \right)}{2!}{{y}^{2}}+\dfrac{n\left( n+1 \right)\left( n+2 \right)}{3!}{{y}^{3}}.............$
If we compare each term of the expansion of ${{\left( 1-y \right)}^{-n}}$ to each term of our equation, we get
$\begin{align}
  & ny=\dfrac{1}{3} \\
 & \Rightarrow y=\dfrac{1}{3n}...............(i) \\
\end{align}$
Also, we get
\[\begin{align}
  & \dfrac{n\left( n+1 \right)}{2}{{y}^{2}}=\dfrac{1.3}{2}{{\left( \dfrac{1}{3} \right)}^{2}} \\
 & \\
\end{align}\]
Now we will substitute the value of y from equation (i). On doing so, we get
\[\begin{align}
  & \dfrac{n\left( n+1 \right)}{2}{{\left( \dfrac{1}{3n} \right)}^{2}}=\dfrac{1.3}{2}{{\left( \dfrac{1}{3} \right)}^{2}} \\
 & \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \dfrac{\left( n+1 \right)}{{{n}^{{}}}}{{\left( \dfrac{1}{3} \right)}^{2}}=\dfrac{1.3}{{}}{{\left( \dfrac{1}{3} \right)}^{2}} \\
 & \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \dfrac{n+1}{n}=3 \\
 & \Rightarrow n=\dfrac{1}{2} \\
 & \\
\end{align}\]
And using the value in equation (i), we can say $y=\dfrac{2}{3}$ .
Therefore, our equation becomes:
$x+1+\dfrac{1}{3}=1+\dfrac{1}{2}\times \dfrac{2}{3}+\dfrac{\dfrac{1}{2}\times \dfrac{3}{2}}{2!}{{\left( \dfrac{2}{3} \right)}^{2}}+\dfrac{\dfrac{1}{2}\times \dfrac{3}{2}\times \dfrac{5}{2}}{3!}{{\left( \dfrac{2}{3} \right)}^{3}}...............$
$\Rightarrow x+1+\dfrac{1}{3}={{\left( 1-\dfrac{2}{3} \right)}^{-\dfrac{1}{2}}}$
$\Rightarrow \dfrac{3x+4}{3}={{\left( \dfrac{1}{3} \right)}^{-\dfrac{1}{2}}}$
Now we will square both sides of our equation. On doing so, we get
\[{{\left( \dfrac{3x+4}{3} \right)}^{2}}={{\left( \dfrac{1}{3} \right)}^{-1}}\]
\[\Rightarrow 9{{x}^{2}}+16+24x={{\left( \dfrac{1}{3} \right)}^{-1}}\times 9\]
\[\Rightarrow 9{{x}^{2}}+16+24x=3\times 9\]
\[\Rightarrow 9{{x}^{2}}+24x=27-16\]
\[\Rightarrow 9{{x}^{2}}+24x=11\]
Hence, we have proved the equation given in the question.

Note: Be careful about the signs and try to keep the equations as neat as possible by removing the removable terms. Moreover, make sure that you learn the formulas related to different binomial expansions as in case of such questions, they are quite useful. It is prescribed that whenever you see a series including terms of the form $^{n}{{C}_{r}}$ or including factorials, then always give a thought of using the binomial expansion as this would give you the answer in the shortest possible manner.