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 $a_{1000}$ :

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

Submit a solution!

Name

Email

School / Institution

Details

Upload solution (*.jpg, *.pdf, *.docx, *.png accepted, max 2mb)

Publish consent

You can display my name on the CMS website if I successfully solve the problem.

Data Protection

I consent to providing this information to Cambridge Maths School. It will be processed in accordance with the Trust Privacy Policy.