Question

Let $a_1, a_2, a_3... a_n (n \in N)$ is a sequence such that $a_1 = a_2 = 2$ and $\frac{2a_{n-1}a_n}{a_{n-1}a_{n+1}-a_n^2} = n^3 - n, \forall n = 2, 3, 4,...$, find the value of $\lfloor\frac{S}{16}\rfloor$, where $S = \sum_{k=2}^{2020} \frac{a_{k+1}}{a_k}$ and $(\lfloor.\rfloor$ denotes G.I.F.)

Answer 1

189

Answer 2

The given recurrence relation is $\frac{2a_{n-1}a_n}{a_{n-1}a_{n+1}-a_n^2} = n^3 - n$.

We can rewrite this as: $\frac{a_{n-1}a_{n+1}-a_n^2}{2a_{n-1}a_n} = \frac{1}{n^3 - n}$

Divide the numerator by $a_{n-1}a_n$: $\frac{a_{n+1}}{2a_n} - \frac{a_n}{2a_{n-1}} = \frac{1}{n(n^2-1)}$ $\frac{a_{n+1}}{2a_n} - \frac{a_n}{2a_{n-1}} = \frac{1}{(n-1)n(n+1)}$

Let $b_n = \frac{a_n}{2a_{n-1}}$. Then the recurrence relation becomes: $b_{n+1} - b_n = \frac{1}{(n-1)n(n+1)}$ for $n = 2, 3, 4, ...$

First, let's find the value of $b_2$. Given $a_1 = a_2 = 2$. $b_2 = \frac{a_2}{2a_1} = \frac{2}{2 \times 2} = \frac{1}{2}$.

Now, we need to find a general expression for $b_N$. We can sum the differences $b_{k+1} - b_k$ from $k=2$ to $N-1$: $\sum_{k=2}^{N-1} (b_{k+1} - b_k) = \sum_{k=2}^{N-1} \frac{1}{(k-1)k(k+1)}$ This is a telescoping sum on the left side: $b_N - b_2$.

For the sum on the right side, we use partial fraction decomposition: $\frac{1}{(k-1)k(k+1)} = \frac{A}{k-1} + \frac{B}{k} + \frac{C}{k+1}$ Multiplying by $(k-1)k(k+1)$: $1 = Ak(k+1) + B(k-1)(k+1) + C(k-1)k$ Setting $k=1 \implies 1 = A(1)(2) \implies A = 1/2$. Setting $k=0 \implies 1 = B(-1)(1) \implies B = -1$. Setting $k=-1 \implies 1 = C(-2)(-1) \implies C = 1/2$. So, $\frac{1}{(k-1)k(k+1)} = \frac{1}{2(k-1)} - \frac{1}{k} + \frac{1}{2(k+1)}$. This can be rewritten as: $\frac{1}{2} \left( \frac{1}{k-1} - \frac{2}{k} + \frac{1}{k+1} \right) = \frac{1}{2} \left[ \left( \frac{1}{k-1} - \frac{1}{k} \right) - \left( \frac{1}{k} - \frac{1}{k+1} \right) \right]$. Let $f(k) = \frac{1}{k-1} - \frac{1}{k}$. Then the term is $\frac{1}{2} [f(k) - f(k+1)]$. The sum is $\sum_{k=2}^{N-1} \frac{1}{2} [f(k) - f(k+1)] = \frac{1}{2} [f(2) - f(N)]$. $f(2) = \frac{1}{1} - \frac{1}{2} = \frac{1}{2}$. $f(N) = \frac{1}{N-1} - \frac{1}{N}$. So, $\sum_{k=2}^{N-1} \frac{1}{(k-1)k(k+1)} = \frac{1}{2} \left[ \frac{1}{2} - \left( \frac{1}{N-1} - \frac{1}{N} \right) \right]$ $= \frac{1}{4} - \frac{1}{2(N-1)N}$.

Now, substitute this back into $b_N - b_2$: $b_N - \frac{1}{2} = \frac{1}{4} - \frac{1}{2N(N-1)}$ $b_N = \frac{1}{2} + \frac{1}{4} - \frac{1}{2N(N-1)}$ $b_N = \frac{3}{4} - \frac{1}{2N(N-1)}$.

We need to find $S = \sum_{k=2}^{2020} \frac{a_{k+1}}{a_k}$. Recall that $b_{k+1} = \frac{a_{k+1}}{2a_k}$, so $\frac{a_{k+1}}{a_k} = 2b_{k+1}$. Using the formula for $b_N$ with $N=k+1$: $2b_{k+1} = 2 \left( \frac{3}{4} - \frac{1}{2(k+1)((k+1)-1)} \right)$ $2b_{k+1} = 2 \left( \frac{3}{4} - \frac{1}{2k(k+1)} \right)$ $\frac{a_{k+1}}{a_k} = \frac{3}{2} - \frac{1}{k(k+1)}$.

Now, substitute this into the sum $S$: $S = \sum_{k=2}^{2020} \left( \frac{3}{2} - \frac{1}{k(k+1)} \right)$. The sum has $2020 - 2 + 1 = 2019$ terms. $S = \sum_{k=2}^{2020} \frac{3}{2} - \sum_{k=2}^{2020} \frac{1}{k(k+1)}$ $S = 2019 \times \frac{3}{2} - \sum_{k=2}^{2020} \left( \frac{1}{k} - \frac{1}{k+1} \right)$.

The first part is $2019 \times \frac{3}{2} = \frac{6057}{2}$. The second sum is a telescoping sum: $\sum_{k=2}^{2020} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{2020} - \frac{1}{2021} \right)$ $= \frac{1}{2} - \frac{1}{2021}$.

Substitute these back into the expression for $S$: $S = \frac{6057}{2} - \left( \frac{1}{2} - \frac{1}{2021} \right)$ $S = \frac{6057}{2} - \frac{1}{2} + \frac{1}{2021}$ $S = \frac{6056}{2} + \frac{1}{2021}$ $S = 3028 + \frac{1}{2021}$.

Finally, we need to find $\lfloor\frac{S}{16}\rfloor$. $\frac{S}{16} = \frac{1}{16} \left( 3028 + \frac{1}{2021} \right)$ $\frac{S}{16} = \frac{3028}{16} + \frac{1}{16 \times 2021}$.

Let's calculate $\frac{3028}{16}$: $3028 \div 16 = 189.25$.

So, $\frac{S}{16} = 189.25 + \frac{1}{16 \times 2021}$. The term $\frac{1}{16 \times 2021}$ is a very small positive number: $16 \times 2021 = 32336$. So, $\frac{1}{16 \times 2021} = \frac{1}{32336} \approx 0.00003$.

Therefore, $\frac{S}{16} = 189.25 + \text{a very small positive fraction}$. $\lfloor\frac{S}{16}\rfloor = \lfloor 189.25 + 0.00003 \rfloor = \lfloor 189.25003 \rfloor = 189$.

The final answer is $\boxed{189}$.

Explanation of the solution:

1. Transform the recurrence relation: The given relation was manipulated to isolate a difference of terms, $b_{n+1} - b_n = \frac{1}{(n-1)n(n+1)}$, by defining $b_n = \frac{a_n}{2a_{n-1}}$.
2. Calculate initial term: $b_2$ was found using $a_1=a_2=2$.
3. Sum the differences: The expression for $b_{n+1}-b_n$ was summed from $n=2$ to $N-1$ to find $b_N$ using a telescoping sum. Partial fraction decomposition was used to simplify the general term of the sum.
4. Express sum terms in $b_k$: The terms of the sum $S = \sum_{k=2}^{2020} \frac{a_{k+1}}{a_k}$ were expressed in terms of $b_{k+1}$ as $\frac{a_{k+1}}{a_k} = 2b_{k+1}$.
5. Evaluate the sum S: The sum $S$ was then evaluated by substituting the derived expression for $2b_{k+1}$ and performing another telescoping sum.
6. Calculate the final value: The value of $S$ was divided by 16, and the floor function was applied to get the final integer result.