Question

The value of $\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right).....\left((2n-1)^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right).....\left((2n)^4+\frac{1}{4}\right)}$ is equal to

• A. $\frac{1}{4n^2+2n+1}$
• B. $\frac{1}{8n^2+4n+1}$
• C. $\frac{1}{4(2n^2+n+1)}$
• D. $\frac{n}{8n^2-4n+1}$

Answer 1

Answer: (B)

$\frac{1}{8n^2+4n+1}$

Answer 2

First, let's factorize the general term $k^4 + \frac{1}{4}$:

$k^4 + \frac{1}{4} = (k^2)^2 + \left(\frac{1}{2}\right)^2$

We can add and subtract $2 \cdot k^2 \cdot \frac{1}{2} = k^2$ to complete the square:

$k^4 + \frac{1}{4} = (k^2)^2 + 2 \cdot k^2 \cdot \frac{1}{2} + \left(\frac{1}{2}\right)^2 - k^2$

$k^4 + \frac{1}{4} = \left(k^2 + \frac{1}{2}\right)^2 - k^2$

This is a difference of squares, $A^2 - B^2 = (A-B)(A+B)$:

$k^4 + \frac{1}{4} = \left(k^2 + \frac{1}{2} - k\right)\left(k^2 + \frac{1}{2} + k\right)$

Rearranging the terms:

$k^4 + \frac{1}{4} = \left(k^2 - k + \frac{1}{2}\right)\left(k^2 + k + \frac{1}{2}\right)$

Let $f(k) = k^2 - k + \frac{1}{2}$.

Notice that $k^2 + k + \frac{1}{2} = (k+1)^2 - (k+1) + \frac{1}{2} = f(k+1)$.

So, the general term can be written as:

$k^4 + \frac{1}{4} = f(k) \cdot f(k+1)$

Now, let's apply this to the given product.

The numerator is:

$N = \left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right).....\left((2n-1)^4+\frac{1}{4}\right)$

Using the factorization:

$N = [f(1)f(2)][f(3)f(4)]...[f(2n-1)f(2n)]$

$N = f(1) \cdot f(2) \cdot f(3) \cdot f(4) \cdot ... \cdot f(2n-1) \cdot f(2n)$

The denominator is:

$D = \left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right).....\left((2n)^4+\frac{1}{4}\right)$

Using the factorization:

$D = [f(2)f(3)][f(4)f(5)]...[f(2n)f(2n+1)]$

$D = f(2) \cdot f(3) \cdot f(4) \cdot f(5) \cdot ... \cdot f(2n) \cdot f(2n+1)$

Now, form the fraction:

$\frac{N}{D} = \frac{f(1) \cdot f(2) \cdot f(3) \cdot f(4) \cdot ... \cdot f(2n-1) \cdot f(2n)}{f(2) \cdot f(3) \cdot f(4) \cdot f(5) \cdot ... \cdot f(2n) \cdot f(2n+1)}$

Most terms cancel out, leaving:

$\frac{N}{D} = \frac{f(1)}{f(2n+1)}$

Now, calculate $f(1)$ and $f(2n+1)$:

$f(k) = k^2 - k + \frac{1}{2}$

$f(1) = 1^2 - 1 + \frac{1}{2} = 1 - 1 + \frac{1}{2} = \frac{1}{2}$

$f(2n+1) = (2n+1)^2 - (2n+1) + \frac{1}{2}$

$f(2n+1) = (4n^2 + 4n + 1) - (2n+1) + \frac{1}{2}$

$f(2n+1) = 4n^2 + 2n + \frac{1}{2}$

Substitute these values back into the expression:

$\frac{N}{D} = \frac{\frac{1}{2}}{4n^2 + 2n + \frac{1}{2}}$

To simplify, multiply the numerator and denominator by 2:

$\frac{N}{D} = \frac{\frac{1}{2} \cdot 2}{\left(4n^2 + 2n + \frac{1}{2}\right) \cdot 2}$

$\frac{N}{D} = \frac{1}{8n^2 + 4n + 1}$