Question

Find the set E of the value of X for which the binomial expansion \[{{\left( 2+5x \right)}^{\dfrac{-1}{2}}}\]is valid.

Answer 1

Hint: Similar to the binomial expansion of \[{{\left( 1+x \right)}^{n}}\], by using the binomial theorem; remove the constant from expression and it should be less than 1.

Complete step-by-step answer:
Binomial expansion is the algebraic expansion of powers of binomials. According the Binomial theorem, it is possible to expand the polynomial \[{{\left( x+y \right)}^{n}}\] into a sum involving terms of the form \[a{{x}^{b}}{{y}^{c}}\], where the exponents b and c are non-negative integer with \[b+c=n\], and the coefficient a of each term is a specific positive integers depending on n and b.

\[\begin{align}
  & \therefore {{\left( x+y \right)}^{n}}=\left( \begin{matrix}
   n \\
   0 \\
\end{matrix} \right){{x}^{n}}{{y}^{0}}+\left( \begin{matrix}
   n \\
   1 \\
\end{matrix} \right){{x}^{n-1}}{{y}^{1}}+\left( \begin{matrix}
   n \\
   2 \\
\end{matrix} \right){{x}^{n-2}}{{y}^{2}}+.....\left( \begin{matrix}
   n \\
   n \\
\end{matrix} \right){{x}^{0}}{{y}^{n}} \\
 & \therefore {{\left( x+y \right)}^{n}}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right){{x}^{n-k}}{{y}^{k}}}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right){{x}^{k}}{{y}^{n-k}}} \\
\end{align}\]
Similarly,
\[\begin{align}
  & {{\left( 1+x \right)}^{n}}=\left( \begin{matrix}
   n \\
   0 \\
\end{matrix} \right){{x}^{0}}+\left( \begin{matrix}
   n \\
   1 \\
\end{matrix} \right){{x}^{1}}+\left( \begin{matrix}
   n \\
   2 \\
\end{matrix} \right){{x}^{2}}+........+\left( \begin{matrix}
   n \\
   n-1 \\
\end{matrix} \right){{x}^{n-1}}+\left( \begin{matrix}
   n \\
   n \\
\end{matrix} \right){{x}^{n}} \\
 & {{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+......+{{x}^{n}} \\
\end{align}\]
\[\therefore {{\left( 1+x \right)}^{n}}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right){{x}^{k}};}\] where, \[\left| x \right|<1\]
 \[\therefore \]In the binomial expansion \[{{\left( 2+5x \right)}^{\dfrac{-1}{2}}}\]can be written as \[{{\left( 2+5x \right)}^{\dfrac{-1}{2}}}\]. Remove the constant term from the binomial expansion.
i.e. \[{{\left[ 2\left( 1+\dfrac{5x}{2} \right) \right]}^{\dfrac{-1}{2}}}={{2}^{\dfrac{-1}{2}}}{{\left( 1+\dfrac{5x}{2} \right)}^{\dfrac{-1}{2}}}\]
Now, \[{{\left( 1+\dfrac{5x}{2} \right)}^{\dfrac{-1}{2}}}\]is similar to \[{{\left( 1+x \right)}^{n}}\]
\[\therefore \left| \dfrac{5x}{2} \right|\]should be less than 1.
\[\begin{align}
  & \Rightarrow \left| \dfrac{5x}{2} \right|<1 \\
 & -1<\dfrac{5x}{2}<1 \\
 & \Rightarrow \dfrac{-2}{5}\end{align}\]
\[\therefore \left( \dfrac{-2}{5},\dfrac{2}{5} \right)\]is the set E of values of x which is valid for the binomial expansion \[{{\left( 2+5x \right)}^{\dfrac{-1}{2}}}\].


Note: here, \[\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right)=\dfrac{n!}{k!\left( n-k \right)!}\]
Where, \[n=0,{{x}^{0}}=1\]and \[\left( \begin{matrix}
   0 \\
   0 \\
\end{matrix} \right)=1\].