Question

Use the fact that nCk equals n!/k!(n-k)! to express in factorials
1) The coefficient "u" of x^n in the expansion of (1+x)^2n
2) the coefficient "v" of x^n in the expansion of (1+x)^2n-1, Hence show that u=2v

Answer 1

Question #74803 Math – Combinatorics – Number Theory

Use the fact that

$$n C k \equiv \frac {n !}{k ! (n - k) !}$$

to express in factorials

1) The coefficient "u" of $x^n$ in the expansion of $(1 + x)^{2n}$

2) The coefficient "v" of $x^n$ in the expansion of $(1 + x)^{2n - 1}$. Hence show that $u = 2v$.

SOLUTION

By the definition,

$$n C k \equiv \frac {n !}{k ! (n - k) !}$$

(More information: https://en.wikipedia.org/wiki/Binomial_coefficient)

By the definition,

$$\begin{array}{l} n! \equiv 1 \cdot 2 \cdot 3 \cdot 4 \cdot \dots \cdot n \\ n! = (n - 1)! \cdot n \\ \end{array}$$

(More information: https://en.wikipedia.org/wiki/Factorial)

Without proof, we assume the validity of formula

$$(x + y)^n = \sum_{k=0}^{n} (n C k) \cdot x^{n-k} y^k$$

(More information: https://en.wikipedia.org/wiki/Binomial_coefficient)

In our case,

1) The coefficient "u" of $x^n$ in the expansion of $(1 + x)^{2n}$

$$(1 + x)^{2n} = \sum_{k=0}^{2n} (2nCk)1^{2n-k}x^k \equiv \sum_{k=0}^{2n} (2nCk)x^k$$

Near the monomial $x^n$ is the coefficient

$$u = 2nCn = \frac{(2n)!}{n! \cdot (2n - n)!} \rightarrow \boxed{u = \frac{(2n)!}{(n!)^2}}$$

2) The coefficient "v" of $x^n$ in the expansion of $(1 + x)^{2n-1}$.

$$(1 + x)^{2n-1} = \sum_{k=0}^{2n-1} ([2n - 1]Ck)1^{(2n-1)-k}x^k \equiv \sum_{k=0}^{2n-1} ([2n - 1]Ck)x^k$$

Near the monomial $x^n$ is the coefficient

$$v = [2n - 1]Cn = \frac{(2n - 1)!}{n! \cdot (2n - 1 - n)!} \rightarrow \boxed{v = \frac{(2n - 1)!}{n! \cdot (n - 1)!}}$$

It remains to show that

$$u = 2v \rightarrow \frac{u}{v} = 2$$

In our case,

$$\frac{u}{v} = \frac{\frac{(2n)!}{(n!)^2}}{\frac{(2n - 1)!}{n! \cdot (n - 1)!}} = \frac{(2n)!}{(n!)^2} \cdot \frac{n! \cdot (n - 1)!}{(2n - 1)!} = \frac{(2n - 1)! \cdot 2n}{n! \cdot n!} \cdot \frac{n! \cdot (n - 1)!}{(2n - 1)!} = \frac{(2n - 1)!}{(2n - 1)!} \cdot \frac{n!}{n!} \cdot \frac{2n \cdot (n - 1)!}{n!} = 1 \cdot 1 \cdot \frac{2n \cdot (n - 1)!}{(n - 1)! \cdot n} = \underbrace{\frac{(n - 1)!}{(n - 1)!}}_{=1} \cdot \frac{n}{\frac{n}{=1}} \cdot \frac{2}{\frac{1}{=2}} = 2$$

Conclusion,

$$\boxed{\frac{u}{v} = 2 \rightarrow u = 2v}$$

ANSWER

$$u = \frac{(2n)!}{(n!)^2}$$ $$v = \frac{(2n-1)!}{n! \cdot (n-1)!}$$ $$u = 2v$$

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