A Recursive Sequence That Hides Its Product

Recursive sequences often tempt us to calculate their terms one by one. In competition problems, that can be exactly the trap.

Sometimes the recurrence itself contains a much simpler structure. Once that structure is exposed, a seemingly complicated product can collapse with almost no computation.

The Problem

Let$$x_1=97,$$

and for $n>1$,$$x_n=\frac{n}{x_{n-1}}.$$

Calculate$$\boxed{\frac{x_1x_2x_3x_4x_5x_6x_7x_8}{8}}.$$

The Hidden Structure: Pairing Adjacent Terms

The recurrence is$$x_n=\frac{n}{x_{n-1}}.$$

Instead of calculating $x_2,x_3,\ldots$ individually, multiply both sides by $x_{n-1}$:$$\boxed{x_{n-1}x_n=n}.$$

This is the key observation.

For example,$$x_1x_2=2,$$$$x_2x_3=3,$$$$x_3x_4=4.$$

We don’t actually need the individual values of the $x_n$‘s.

The recurrence has already given us the products we need.

Pair the Terms

Let$$P=x_1x_2x_3x_4x_5x_6x_7x_8.$$

Group the terms into adjacent pairs:$$P=(x_1x_2)(x_3x_4)(x_5x_6)(x_7x_8).$$

Now apply$$x_{n-1}x_n=n$$

to each pair:$$x_1x_2=2,$$$$x_3x_4=4,$$$$x_5x_6=6,$$$$x_7x_8=8.$$

Therefore,$$P=2\cdot4\cdot6\cdot8.$$

Hence,$$P=384.$$

The required expression is$$\frac{P}{8}=\frac{384}{8}=\boxed{48}.$$

Why Was $x_1=97$ Given?

This is an important part of the problem.

The value $97$ looks significant because it determines the actual sequence:

and so on.

But none of those individual terms needs to be calculated.

The number $97$ is deliberate misdirection.

The requested product depends only on the paired relationships

$x_1x_2=2, \qquad x_3x_4=4, \qquad x_5x_6=6, \qquad x_7x_8=8$

In fact, the same answer results for any nonzero starting value $x_1$.

The General Pattern

Suppose$$x_n=\frac{n}{x_{n-1}}.$$

Then$$x_{n-1}x_n=n.$$

For an even number $2m$ of terms,$$x_1x_2\cdots x_{2m} =(x_1x_2)(x_3x_4)\cdots(x_{2m-1}x_{2m}).$$

Therefore,$$x_1x_2\cdots x_{2m}=2\cdot4\cdot6\cdots2m.$$

Since$$2\cdot4\cdot6\cdots2m=2^m m!,$$

we obtain$$\boxed{x_1x_2\cdots x_{2m}=2^m m!}.$$

The individual terms may look complicated, but their product has a remarkably simple form.

The Competition Insight

When a recursive sequence appears inside a product, don’t automatically calculate the terms.

First ask:

Can the recurrence be rearranged into a useful relationship between consecutive terms?

Here,$$x_n=\frac{n}{x_{n-1}}$$

immediately becomes$$x_{n-1}x_n=n.$$

That turns a recursive problem into a simple pairing problem.

The Takeaway

Don’t compute the sequence when the recurrence already gives you the product.

The structure is often more valuable than the arithmetic.