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Math  /  Algebra

Question8. (a) Verify that for all n≥1n \geq 1, 2⋅6⋅10⋅14⋯(4n−2)=(2n)!n!2 \cdot 6 \cdot 10 \cdot 14 \cdots(4 n-2)=\frac{(2 n)!}{n!} (b) Use part (a) to obtain the inequality 2n(n!)2≤(2n)2^{n}(n!)^{2} \leq(2 n) ! for all n≥1n \geq 1.

Studdy Solution

STEP 1

1. We are given a product sequence and a factorial expression to verify for all n≥1 n \geq 1 .
2. We will use mathematical induction for verification.
3. We will use algebraic manipulation and properties of inequalities to derive the inequality in part (b).

STEP 2

1. Verify the given equation using mathematical induction.
2. Use the verified equation to derive the inequality.

STEP 3

First, verify the base case for n=1 n = 1 .
For n=1 n = 1 , the left-hand side of the equation is:
2⋅6⋅10⋅14⋯(4⋅1−2)=2 2 \cdot 6 \cdot 10 \cdot 14 \cdots (4 \cdot 1 - 2) = 2
The right-hand side of the equation is:
(2⋅1)!1!=2!1!=21=2 \frac{(2 \cdot 1)!}{1!} = \frac{2!}{1!} = \frac{2}{1} = 2
Since both sides are equal, the base case holds true.

STEP 4

Assume the statement is true for some n=k n = k , i.e.,
2⋅6⋅10⋅14⋯(4k−2)=(2k)!k! 2 \cdot 6 \cdot 10 \cdot 14 \cdots (4k - 2) = \frac{(2k)!}{k!}
Now, prove it for n=k+1 n = k + 1 .
The left-hand side for n=k+1 n = k + 1 is:
2⋅6⋅10⋅14⋯(4k−2)⋅(4(k+1)−2) 2 \cdot 6 \cdot 10 \cdot 14 \cdots (4k - 2) \cdot (4(k + 1) - 2)
Using the induction hypothesis, this becomes:
(2k)!k!⋅(4k+2) \frac{(2k)!}{k!} \cdot (4k + 2)
The right-hand side for n=k+1 n = k + 1 is:
(2(k+1))!(k+1)!=(2k+2)!(k+1)! \frac{(2(k + 1))!}{(k + 1)!} = \frac{(2k + 2)!}{(k + 1)!}
We need to show:
(2k)!k!⋅(4k+2)=(2k+2)!(k+1)! \frac{(2k)!}{k!} \cdot (4k + 2) = \frac{(2k + 2)!}{(k + 1)!}
Simplifying the right-hand side:
(2k+2)(2k+1)(2k)!(k+1)k! \frac{(2k + 2)(2k + 1)(2k)!}{(k + 1)k!}
Both expressions are equal, thus proving the induction step.

STEP 5

Now, use the verified equation to derive the inequality.
Given:
2⋅6⋅10⋅14⋯(4n−2)=(2n)!n! 2 \cdot 6 \cdot 10 \cdot 14 \cdots (4n - 2) = \frac{(2n)!}{n!}
We can rewrite the left-hand side as:
2n⋅n! 2^n \cdot n!
Thus, we have:
2n⋅n!=(2n)!n! 2^n \cdot n! = \frac{(2n)!}{n!}
Multiplying both sides by n! n! , we get:
2n⋅(n!)2=(2n)! 2^n \cdot (n!)^2 = (2n)!
This implies:
2n⋅(n!)2≤(2n)! 2^n \cdot (n!)^2 \leq (2n)!
The inequality is:
2n(n!)2≤(2n)! 2^n (n!)^2 \leq (2n)!

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