Attractive

Calculus Level 5

$\large \frac{\left(1^4 + \frac14\right)\left(3^4 + \frac14\right)\cdots \left((2n-1)^4 + \frac14\right)}{\left(2^4 + \frac14\right)\left(4^4 + \frac14\right)\cdots \left((2n)^4 + \frac14\right)}$

Given that the expression above can be simplified to $(an^2+ bn + c)^{-1}$ for constants $a,b,c$ , find $21(a-b+c)$ .

The answer is 105.

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Rindell Mabunga
Dec 15, 2014

Just to improve Sir Martin Sergio's Solution :)

Let

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

Completing the squares,

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

But this is clearly difference of two squares,

$f(n) = \frac{\prod_{i = 1}^n [(((2i - 1)^2 + \frac{1}{2})-(2i - 1))(((2i - 1)^2 + \frac{1}{2})+(2i - 1))]}{\prod_{i = 1}^n [(((2i)^2 + \frac{1}{2})-(2i))(((2i)^2 + \frac{1}{2})+(2i))]}$

The portion

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

But the portion

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

Therefore the two factors will cancel leading to,

$f(n) = \frac{\prod_{i = 1}^n [((2i - 1)^2 + \frac{1}{2})-(2i - 1))]}{\prod_{i = 1}^n [((2i)^2 + \frac{1}{2})+(2i)]}$

Similarly, other factors will also cancel leading to

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

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

$f(n) = \frac{1}{8n^2 + 4n + 1}$

$a = 8$

$b = 4$

$c = 1$

$21(a - b + c) = 21(8 - 4 + 1) = 21(5) = \boxed{105}$

Hi, that's an elegant solution! :-) Didn't consider that approach yesterday. Thumbs up to you.

Martin Sergio H. Faester - 6 years, 5 months ago

Actually, your solution was a short step away. Assuming that the answer must have the form of $\frac{ 1}{ an^2 + bn + c }$ , you showed that it must be $\frac{ 1}{ 8n^2 + 4n + 1 }$ .

If so, this implies that $\frac{ ( 2n -1 ) ^ 4 + \frac{1}{4} } { (2n)^4 + \frac{1}{4} } = \frac{ 8 (n-1) ^2 + 4 (n-1) + 1} { 8n^2 + 4n + 1 }$ . This helps us to motivate the telescoping product, instead of it being something that is mysteriously obtained.

Of course, if we do not know the form of the answer, then we must have some speculative guesses about what it is like. A priori, there is no reason why it must be so nice (other than the fact that it is set as a question).

Calvin Lin Staff - 6 years, 5 months ago

Martin Sergio H. Faester
Dec 15, 2014

Let
$f(n) = \frac{\prod_{i=1}^n \left((2i-1)^4+\frac{1}{4}\right)}{\prod_{i=1}^n \left((2i)^4+\frac{1}{4}\right)}$

The exercise states that $f(n) = \frac{1}{a n^2 + b n + c}$ , so we can solve the problem by computing $f(1), f(2)$ and $f(3)$ , and solve the emerging system of linear equations. Since $f(1) = \frac{1}{13}, f(2) = \frac{1}{41}$ and $f(3) = \frac{1}{85}$ , we get

$\begin{aligned} \frac{1}{f(1)} & = a + b + c = 13 \\ \frac{1}{f(2)} & = 4a + 2b + c = 41 \\ \frac{1}{f(3)} & = 9a + 3b + c = 85 \end{aligned}$

The solution for this system of linear equations is $a = 8, b = 4$ and $c = 1$ . Hence, $21( a - b + c) = 21 \cdot 5 = \boxed{105}$

Now that you "know" we must have $a = 8, b = 4 , c = 1$ , can you prove it for all $n$ ?

Hint: Telescoping Product.

Calvin Lin Staff - 6 years, 6 months ago

For proving , it would be easier to take the help of PMI , is there any other way?

U Z - 6 years, 5 months ago

Attractive

The answer is 105.

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