Question

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

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.

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 $\left( \begin{matrix} 0 \\ 0 \\ \end{matrix} \right)=1$.