Math  /  Algebra

Question8. (a) Verify that for all n1n \geq 1, 261014(4n2)=(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 n1n \geq 1.

Studdy Solution

STEP 1

1. We are given a product sequence and need to verify its equivalence to a factorial expression.
2. The sequence is an arithmetic sequence with a common difference of 4.
3. We will use the result from part (a) to derive an inequality in part (b).

STEP 2

1. Verify the equality for the product sequence and factorial expression.
2. Use the verified equality to derive the inequality.

STEP 3

Verify the equality for n=1 n = 1 as a base case:
For n=1 n = 1 , the left-hand side (LHS) is: 2610(412)=2 2 \cdot 6 \cdot 10 \cdots (4 \cdot 1 - 2) = 2
The right-hand side (RHS) is: (21)!1!=2!1!=21=2 \frac{(2 \cdot 1)!}{1!} = \frac{2!}{1!} = \frac{2}{1} = 2
Since LHS = RHS, the base case holds.

STEP 4

Assume the equality holds for some n=k n = k , i.e., 2610(4k2)=(2k)!k! 2 \cdot 6 \cdot 10 \cdots (4k - 2) = \frac{(2k)!}{k!}
Now, verify for n=k+1 n = k + 1 :
The LHS for n=k+1 n = k + 1 is: 2610(4k2)(4(k+1)2) 2 \cdot 6 \cdot 10 \cdots (4k - 2) \cdot (4(k + 1) - 2)
By the induction hypothesis: =(2k)!k!(4k+2) = \frac{(2k)!}{k!} \cdot (4k + 2)
The RHS 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)!}
Expanding the RHS: (2k+2)(2k+1)(2k)!(k+1)k! \frac{(2k + 2)(2k + 1)(2k)!}{(k + 1)k!}
Simplifying, we see both sides are equal, thus the induction step holds.

STEP 5

Using the verified equality, derive the inequality:
From part (a), we have: 2610(4n2)=(2n)!n! 2 \cdot 6 \cdot 10 \cdots (4n - 2) = \frac{(2n)!}{n!}
Notice: 2610(4n2)=2nn! 2 \cdot 6 \cdot 10 \cdots (4n - 2) = 2^n \cdot n!
Thus: 2nn!=(2n)!n! 2^n \cdot n! = \frac{(2n)!}{n!}
Rearranging gives: 2n(n!)2(2n)! 2^n (n!)^2 \leq (2n)!
The inequality is:
2n(n!)2(2n)! 2^n (n!)^2 \leq (2n)!

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