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Problem of the Week- Week 5 of COMC

Can anyone please tell me if my solution is correct.

So, I divide the equation by "$a_n$" in the numerator and denominator ($a_n$ cannot be zero).

I get: $a_{n+1} =$ 1 / (n + 1/(a_n))

Using this new equation, I find $a_{n + 1}$ in terms of $a_{n - 1}$, then $a_{n -2}$ and so on. I see the pattern is

$a_{n + 1} =$ 1 / (sum of numbers k to n + (1 / a_k))

I substitute $n + 1 = 1996$, $n = 1995$, $k = 2$, and solve for $a_2$.

Then I get $a_{1996}$ is $\frac{1}{1991011}$

Shraddheya Shendre
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Alan Tam
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1 Answers1

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If $a_{n+1} =\dfrac1{n+\frac1{a_n}} $ then $\dfrac1{a_{n+1}} =n+\dfrac1{a_n} $.

Letting $b_n =\dfrac1{a_n} $, $b_{n+1} =n+b_n $, or $b_{n+1}-b_n = n$.

Summing from $1$ to $m-1$, $b_m-b_1 =\sum_{n=1}^{m-1}(b_{n+1}-b_n) =\sum_{n=1}^{m-1}n =m(m-1)/2 $, so $b_m =1+m(m-1)/2 =(m^2-m+2)/2 $ so that $a_m =\dfrac{2}{m^2-m+2} $.

If $m = 1996$, $(m^2-m+2)/2 =1991011 $, so I agree.

marty cohen
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