4. (i) A sequence $U_1, U_2, U_3, ...$ is defined by $U_{n+2} = 2U_{n+1} - U_n$, $n \geq 1$ - Edexcel - A-Level Maths Pure - Question 6 - 2015 - Paper 1

First, we need to calculate the terms of the sequence up to $U_{20}$ using the recurrence relation:

1. $U_1 = 4$
2. $U_2 = 4$
3. $U_3 = 4$
4. $U_4 = 2U_3 - U_2 = 2(4) - 4 = 4$
5. Continuing this way, we can calculate:
   • $U_5 = 4$
   • $U_6 = 4$
   • ...

In fact, we observe that $U_n = 4$ for all $n$ from 1 to 20.

Thus, we have:

$\sum_{n=1}^{20} U_n = 20 \times 4 = 80$

To find $V_n$, we again use the recurrence relation:

$V_n = 2V_{n-1} - V_{n-2}$

Starting with the given initial conditions:

• $V_1 = k$
• $V_2 = 2k$

Calculating subsequent terms:

1. $V_3 = 2V_2 - V_1 = 2(2k) - k = 4k - k = 3k$
2. $V_4 = 2V_3 - V_2 = 2(3k) - 2k = 6k - 2k = 4k$
3. $V_5 = 2V_4 - V_3 = 2(4k) - 3k = 8k - 3k = 5k$

This suggests a pattern, which leads us to the formula: $V_n = (n + 1)k$

We first calculate:

$\sum_{n=1}^{5} V_n = V_1 + V_2 + V_3 + V_4 + V_5$ Substituting our expressions for each:

$\sum_{n=1}^{5} V_n = k + 2k + 3k + 4k + 5k = (1 + 2 + 3 + 4 + 5)k = 15k$

Setting this equal to 165 gives:

$15k = 165$

Now, solving for $k$:

$k = \frac{165}{15} = 11$