Question

(a) Show that if$\mathit{n}$is a positive integer, then $\left(\begin{array}{c}\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ n\end{array}\right)\mathbf{=}\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)}{{\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{)}}^{\mathbf{n}}}$

(b) Use the extended binomial theorem and part (a) to show that the coefficient of ${\mathit{x}}^{\mathbf{n}}$in the expansion of$\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{4}\mathit{x}{\mathbf{)}}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}$is$\left[\begin{array}{c}2n\\ n\end{array}\right]$for all nonnegative integers$\mathit{n}$

Short answer

(a) The resultant answer is$\left(\begin{array}{c}\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ n\end{array}\right)=\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)}{{(-4)}^n}$

(b) The coefficient of$x^n$ is$\left(\begin{array}{c}2n\\ n\end{array}\right)$

Step-by-step solution

Step 1:  Given data

The given data$n$ is a positive integer.

Step 2: Concept of Extended binomial theorem

Generating function for the sequence ${\mathit{a}}_{\mathbf{0}}\mathbf{,}{\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{a}}_{\mathbf{k}}\mathbf{,}\mathbf{\dots }$ of real numbers is the infiniteseries:

$\mathit{G}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{a}}_{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}{\mathit{a}}_{\mathbf{k}}{\mathit{x}}^{\mathbf{k}}\mathbf{+}\mathbf{\dots }\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{+}\mathbf{\infty }}{\mathit{a}}_{\mathbf{k}}{\mathit{x}}^{\mathbf{k}}$

Extended binomial theorem:

${\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}}^{\mathbf{u}}\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{+}\mathbf{\infty }}\left(\begin{array}{c}u\\ k\end{array}\right){\mathit{x}}^{\mathbf{k}}$

 Step 3:Simplify by using the concept of Binomial theorem

(a)

Use the binomial theorem to simplify the expression;

$\left(\begin{array}{c}-1/2\\ n\end{array}\right)=\frac{(-1/2)(-1/2-1)\dots (-1/2-n+1)}{n!}$

$\left(\begin{array}{c}-1/2\\ n\end{array}\right)=\frac{(-1/2)(-3/2)\dots (1/2-n)}{n!}$

$\left(\begin{array}{c}-1/2\\ n\end{array}\right)={(-1)}^n·\frac{(1/2)(3/2)\dots (n-1/2)}{n!}$

$\left(\begin{array}{c}-1/2\\ n\end{array}\right)=\frac{{(-1)}^n}{2^n}·\frac{1·3·5·\dots ·(2n-1)}{n!}$

Similarly; use the binomial theorem to simplify the expression

$\left(\begin{array}{c}-1/2\\ n\\ \end{array}\right)=\frac{{(-1)}^n}{2^n}·\frac{1·3·5·\dots ·(2n-1)}{n!}·\frac{2^{n}n!}{2^{n}n!}$

localid="1668673173414" $\left(\begin{array}{c}-1/2\\ n\end{array}\right)=\frac{{(-1)}^n}{2^n}·\frac{1·3·5·\dots ·(2n-1)}{n!}·\frac{2^n(1·2·3·\dots ·n)}{2^{n}n!}$

localid="1668673188900" $\left(\begin{array}{c}-1/2\\ n\end{array}\right)=\frac{{(-1)}^n}{2^n}·\frac{1·3·5·\dots ·(2n-1)}{n!}·\frac{2·4·6·\dots ·2n}{2^{n}n!}$

localid="1668673200889" $\left(\begin{array}{c}-1/2\\ n\end{array}\right)=\frac{{(-1)}^n\left(2n\right)!}{2^n{2}^{n}n!n!}$

Similarly; use the binomial theorem to simplify the expression

localid="1668673026604" $\left(\begin{array}{c}-1/2\\ n\end{array}\right)=\frac{{(-1)}^n}{4^n}\left(\begin{array}{c}2n\\ n\end{array}\right)$

localid="1668672975583" $\left(\begin{array}{c}-1/2\\ n\end{array}\right)=\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)}{{(-4)}^n}$

Hence, the resultant answer is

localid="1668673296570" $\left(\begin{array}{c}-1/2\\ n\end{array}\right)=\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)}{{(-4)}^n}$

Use the binomial theorem and simplify the expression

(b)

Use the extended binomial theorem:

${(1-4x)}^{-1/2}={(1+(-4x\left)\right)}^{-1/2}$

localid="1668673345900" ${(1-4x)}^{-1/2}=\sum _{k=0}^{+\infty }\left(\begin{array}{c}-1/2\\ k\end{array}\right){(-4x)}^k$

localid="1668673373014" ${(1-4x)}^{-1/2}=\sum _{k=0}^{+\infty }\frac{\left(\begin{array}{c}2k\\ k\end{array}\right)}{{(-4)}^k}{(-4)}^k{x}^k\text{Use part (a)}$

localid="1668673399092" ${(1-4x)}^{-1/2}=\sum _{k=0}^{+\infty }\left(\begin{array}{c}2k\\ k\end{array}\right)x^k$

Thus, note that the coefficient of localid="1668673523762" $x^n$is$\left(\begin{array}{c}2n\\ n\end{array}\right)$