4. (i) Show that $$ \sum_{r=1}^{16} (3 + 5r + 2^r) = 131798 $$ (ii) A sequence $u_n, u_1, u_2, ...$ is defined by $$ u_{n+1} = \frac{1}{u_n}, \quad u_1 = \frac{2}{3} $$ Find the exact value of $$ \sum_{r=1}^{100} u_r $$ - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 2

To solve the summation, we can break it down into three parts:

1. Constant term: $\sum_{r=1}^{16} 3 = 3 \times 16 = 48$

2. Linear term: The sum of an arithmetic series is given by: $\sum_{r=1}^{n} ar = \frac{n}{2}(first\ term + last\ term)$ Here, the first term (for r=1) is 5 and the last term (for r=16) is 5x16 = 80. Thus: $\sum_{r=1}^{16} 5r = 5 \times \frac{16}{2}(1 + 16) = 5 \times 8 \times 17 = 680$

3. Exponential term: Applying the formula for the sum of a geometric progression: $\sum_{r=1}^{n} ar^r = a \frac{r(1 - r^n)}{1 - r}$, where a=2 and r=2. Thus: $\sum_{r=1}^{16} 2^r = 2 \frac{1 - 2^{16}}{1 - 2} = 2(2^{16} - 1) = 2^{17} - 2$ Computing $2^{17}$ gives 131072, hence: $\sum_{r=1}^{16} 2^r = 131072 - 2 = 131070$

Combining the results: $48 + 680 + 131070 = 131798$ This confirms that $\sum_{r=1}^{16} (3 + 5r + 2^r) = 131798$.