Question

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

Answer 1

Hint: Similar to the binomial expansion of $${(1+x)}^n$$, 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 $${(x+y)}^n$$ 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.

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


Note: here, $$\left(\begin{array}{c}n\\ k\end{array}\right)={\displaystyle \frac{n!}{k!(n-k)!}}$$
Where, $$n=0,x^0=1$$and $$\left(\begin{array}{c}0\\ 0\end{array}\right)=1$$.