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.

& \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{aligned}$$ Similarly, $$\begin{aligned} & {{\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{aligned}$$ $$\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{aligned} & \Rightarrow \left| \dfrac{5x}{2} \right|<1 \\\ & -1<\dfrac{5x}{2}<1 \\\ & \Rightarrow \dfrac{-2}{5}\end{aligned}$$ $$\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$$.