A sequence of numbers $a_1, a_2, a_3, \ldots$ is defined by a_{n+1} = \frac{k(a_n + 2)}{a_n}, \quad n \in \mathbb{N} where k is a constant - Edexcel - A-Level Maths Pure - Question 15 - 2020 - Paper 1

If $k = 1$, the sequence becomes:

$a_{n+1} = \frac{a_n + 2}{a_n}$

Thus, all terms would be equal and would not allow the sequence to have a period of 3. This would mean:

1. If $a_1 = 2$, then:
   • $a_2 = \frac{2 + 2}{2} = 2$,
   • $a_3 = 2$,
   • $a_4 = 2$, and so on.

Hence, all terms will be the same and thus cannot form a periodic sequence of order 3. Therefore, we conclude that $k \neq 1$.

To find this sum, first we identify the repeating terms in the sequence:

We have established that:

• $a_1 = 2$,
• $a_2 = 2k$,
• $a_3 = k + 1$.

Now, using $k=2$ from the equation earlier, we find:

• $a_1 = 2$,
• $a_2 = 2 \cdot 2 = 4$,
• $a_3 = 2 + 1 = 3$.

The sequence repeats every 3 terms as: 2, 4, 3.

Next, calculate the total number of complete cycles in 80 terms:

• Each cycle contains 3 terms, thus: $\frac{80}{3} = 26 \text{ complete cycles with a remainder of } 2$

The total contribution of the 26 complete cycles is:

$26 \cdot (2 + 4 + 3) = 26 \cdot 9 = 234$

For the remaining 2 terms:

• They contribute: $2 + 4 = 6$.

Therefore, the total value is: $\sum_{n=1}^{80} a_n = 234 + 6 = 240$