British Maths Olympiad 1997 Round 1 Problem 2

My solution to an Olympiad problem.

By Vamshi Jandhyala in mathematics

September 25, 2019

For positive integers $n$, the sequence $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ is defined by

$$ \begin{align*} a_{1} = 1; a_{n} = (\frac{n+1}{n-1})(a_{1}+ a_{2}+ a_{3}+ \dots + a_{n-1}), n>1 \end{align*} $$

Determine the value of $a_{1997}$.

Solution

Solution 1

We have

$$ \begin{align*} a_{n+1} &=& a_{n}\frac{2(n+2)}{n+1} \
\implies \frac{a_{2}}{a_{1}}\frac{a_{3}}{a_{2}}\cdots\frac{a_{n}}{a_{n-1}} = a_{n} &=& 2^{n-2}(n+1) \end{align*} $$

Solution 2 (Brute force computational solution)

n = 5
sumfn = 1
for i in range(2, n+1):
  fn_next = (sumfn*(i+1))/(i-1)
  sumfn += fn_next
print("f(%d) = %s" % (n,fn_next))