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Math  /  Algebra

Question(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 \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 need to verify its equivalence to a factorial expression.
2. We will use mathematical induction to verify the equality for all n≥1 n \geq 1 .
3. We will use the verified equality to derive an inequality.

STEP 2

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

STEP 3

To verify the equality, start with the base case n=1 n = 1 .
For n=1 n = 1 , the left-hand side is: 2⋅(4⋅1−2)=2 2 \cdot (4 \cdot 1 - 2) = 2
The right-hand side is: (2⋅1)!1!=2!1!=21=2 \frac{(2 \cdot 1)!}{1!} = \frac{2!}{1!} = \frac{2}{1} = 2
Both sides are equal, so the base case holds.

STEP 4

Assume the equality holds for n=k n = k , i.e., 2⋅6⋅10⋅⋯⋅(4k−2)=(2k)!k! 2 \cdot 6 \cdot 10 \cdot \cdots \cdot (4k - 2) = \frac{(2k)!}{k!}
We need to show it holds for n=k+1 n = k + 1 .
The left-hand side for n=k+1 n = k + 1 is: 2⋅6⋅10⋅⋯⋅(4k−2)⋅(4(k+1)−2) 2 \cdot 6 \cdot 10 \cdot \cdots \cdot (4k - 2) \cdot (4(k+1) - 2)
By the induction hypothesis, this equals: (2k)!k!⋅(4k+2) \frac{(2k)!}{k!} \cdot (4k + 2)

STEP 5

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)!}
Notice that: (2k+2)!=(2k+2)(2k+1)(2k)! (2k + 2)! = (2k + 2)(2k + 1)(2k)!
Thus: (2k+2)!(k+1)!=(2k+2)(2k+1)(2k)!(k+1)k! \frac{(2k + 2)!}{(k+1)!} = \frac{(2k + 2)(2k + 1)(2k)!}{(k+1)k!}
Simplifying, we get: (2k)!k!⋅(4k+2)=(2k+2)(2k+1)(2k)!(k+1)k! \frac{(2k)!}{k!} \cdot (4k + 2) = \frac{(2k + 2)(2k + 1)(2k)!}{(k+1)k!}
This confirms the induction step, completing the proof.

STEP 6

Using the verified equality, we have: 2⋅6⋅10⋅⋯⋅(4n−2)=(2n)!n! 2 \cdot 6 \cdot 10 \cdot \cdots \cdot (4n - 2) = \frac{(2n)!}{n!}
This implies: 2n⋅n!=(2n)!n! 2^n \cdot n! = \frac{(2n)!}{n!}
Rearranging gives: 2n⋅(n!)2≤(2n)! 2^n \cdot (n!)^2 \leq (2n)!
The inequality derived is:
2n(n!)2≤(2n)! 2^n (n!)^2 \leq (2n)!

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