Looks like Wallis to me

Algebra Level 2

Given that $f(x) = \frac {x^2}{x^2 - 1}$ , compute

$51\times \big[ f(50)\times f(49)\times \cdots \times f(3)\times f(2) \big].$

The answer is 100.

1 solution

Rmflute Shrivastav
Mar 29, 2015

We start with a smaller group of Numbers. f(50) * f(49) * f(48)

Simplifying, we get.... (We factored the denominator of f(x))

$\frac {50 ^ 2} {(50 + 1)(50-1)} * \frac {49 ^ 2} {(49+ 1)(49-1)} * \frac {48 ^ 2} {(48+ 1)(48-1)}$

We can see that we can cancel some terms. After cancelling we get....

$\frac {50 * 48} {51 * 47}$

WE see that on the top we get the first number and the last number multiplied together, while on the bottom we get the first number + 1 and the last number -1.

Using the generalization we get...

$\frac {50 * 2} {51 * 1}$

We have to multiply the overall expression by 51, so we can simplify to 100. That is the answer.

(If you have any question please ask below, I shortened my explanation because I do not know how to use Latex and it was getting hard to write the solution using the new Latex Method that I just learned (kind of))

Here's a way to formally write the solution using product notation (which involves the idea of your solution):

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

The expression (call it $\mathcal{P}$ ) given to us is:

$\mathcal{P}=51\cdot\prod_{x=2}^{50}f(x)=51\cdot\prod_{x=2}^{50}\left(\frac{x}{x-1}\cdot\frac{x}{x+1}\right)\\ \frac{\mathcal{P}}{51}=\left(\frac{\displaystyle\left(\prod_{x=2}^{50} x\right)^2}{\left(\displaystyle\prod_{x=1}^{49}x\right)\cdot\left(\displaystyle\prod_{x=3}^{51}x\right)}\right)=\frac{2\times 50}{51}\\ \implies \frac{\mathcal{P}}{51}=\frac{100}{51}\implies \boxed{\mathcal{P}=100}$

A few notations used:

$\quad\displaystyle\prod_{k=a}^bk=a(a+1)(a+2)\cdots (b-1)b$

Prasun Biswas - 6 years, 1 month ago

Looks like Wallis to me

The answer is 100.

1 solution

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