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Problem of the Week

Problem 10 solution

Congratulations to Andrei for solving Problem 10.

We may initially rewrite the given equation as:

    \[ a_{n+1} = \frac{2n+2}{2n+1} \cdot a_n \]

    \[ \Rightarrow a_n = \frac{2(n-1) + 2}{2(n-1) + 1} \cdot a_{n-1} = \frac{2n}{2n-1} \cdot a_{n-1} \]

We can express a_{n+1} as the product:

    \[ a_{n+1} = \left(\frac{2n+2}{2n+1}\right) \cdot \left(\frac{2n}{2n-1}\right) \cdots a_0 \]

Now, let’s consider this large fraction in separate parts, namely the numerator and the denominator.

Numerator:

The numerator is the product of all even numbers from 2n+2 to 2 (decreasing by 2 each time).

This can be expressed as:

    \[ 2 \cdot 4 \cdots (2n+2) = 2^{n+1} \cdot (n+1)! \]

This is derived by dividing each term in the series 2 \cdot 4 \cdots (2n+2) by 2, which is done n+1 times, resulting in 2^{n+1}. The remaining terms form (n+1)!.

Hence, the numerator equals 2^{n+1} \cdot (n+1)!.

Denominator:

The denominator is the product of all odd numbers from 2n+1 to 1.

To calculate this, we consider (2n+2)!, which equals 1 \cdot 2 \cdot 3 \cdots (2n+1) \cdot (2n+2).

By dividing (2n+2)! by the product of all even terms, we isolate the odd terms, resulting in:

    \[ \frac{(2n+2)!}{2^{n+1} \cdot (n+1)!} \]

Therefore, the fraction becomes:

    \[ \frac{(2^{n+1} \cdot (n+1)!)^2}{(2n+2)!} \]

Using this, we find:

    \[ a_{n+1} = \frac{(2^{n+1} \cdot (n+1)!)^2}{(2n+2)!} \]

    \[ \Rightarrow a_n = \frac{(2^n \cdot (n)!)^2}{(2n)!} \]

Finally, for a_{1000}:

    \[ a_{1000} = \frac{(2^{1000} \cdot 1000!)^2}{2000!} \]

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