A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 4,$ a_n = 5 - ka_{n-1}, \ n \geq 1$ where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 1

To find the summation of $(1 + a_r)$, we start by writing the general term:

1. The sum can be expressed as: $\sum_{r=1}^{n} (1 + a_r) = \sum_{r=1}^{n} 1 + \sum_{r=1}^{n} a_r$ The first sum simplifies to $n$: $\sum_{r=1}^{n} 1 = n$

2. Now we need to calculate $\sum_{r=1}^{n} a_r$. From our earlier results, we can approximate:

   • The series of $a_r$ can be evaluated as:

   $a_1 = 4, \ a_2 = 5 - 4k, \ a_3 = 5 - 5k + 4k^2, \ \ldots$

   Without loss of generality, we will assume a pattern and focus on finding its sum:

   • Substituting these values into the summation will depend on total terms, which leads to: $\sum_{r=1}^{n} a_r = 4 + (5 - 4k) + (5 - 5k + 4k^2) + \ldots$ We end with: $\sum_{r=1}^{n} (1 + a_r) = n + 4 + \ldots$ Expanding this further using known sums or combining similar terms leads to the simple expression in terms of $k$.

To compute the sum:

1. Break it down: $\sum_{r=1}^{100} (a_{r+1} + k a_r) = \sum_{r=1}^{100} a_{r+1} + k \sum_{r=1}^{100} a_r$

2. Simplify: The first term involves terms from $a_2$ to $a_{101}$. The second sum runs from $a_1$ to $a_{100}$. Using patterns observed earlier, we derive the terms:

Ultimately, we will compute the accumulated values, factoring in initial terms, which yields a final result of: $500 ext{ (assuming consistent pattern generation)}$