Summation and series: DHS 2023 Promo Q7

(a)

(i) $$\begin{aligned} \sum_{r=1}^{n}\left(2^{r+1} + 3r - r^2\right) &= 2\sum_{r=1}^{n} 2^r + 3\sum_{r=1}^{n} r - \sum_{r=1}^{n} r^2\\ &= 2\left[\frac{2(2^n - 1)}{2 - 1}\right] + 3\left[\frac{n}{2}(n+1)\right] - \frac{1}{6}n(n+1)(2n+1)\\ &= 4(2^n - 1) + \frac{n(n+1)}{6}\big[9 - (2n+1)\big]\\ &= 4(2^n - 1) + \frac{1}{3}n(n+1)(4 - n). \end{aligned}$$

(a)

(ii) Replace $r$ with $r+1$. $$\begin{aligned} \sum_{r=4}^{N}\left(2^r + 3r - (r-1)^2\right) &= \sum_{r=3}^{N-1}\left(2^{r+1} + 3(r+1) - r^2\right)\\ &= \sum_{r=3}^{N-1}\left(2^{r+1} + 3r - r^2\right) + \sum_{r=3}^{N-1} 3\\ &= \sum_{r=1}^{N-1}\left(2^{r+1} + 3r - r^2\right) - \sum_{r=1}^{2}\left(2^{r+1} + 3r - r^2\right) + 3(N - 3)\\ &= 4\!\left(2^{N-1} - 1\right) + \frac{1}{3}N(N-1)(5 - N) - 16 + (3N - 9)\\ &= 2\!\left(2^N\right) + \frac{1}{3}N(N-1)(5 - N) + 3N - 29, \end{aligned}$$ so $B = 2$, $C = \dfrac{1}{3}$, $D = 3$ and $E = -29$.

(b)

(i) $p=5$

Method 1 (GC) (preferred)

Method 2 (algebraic) $$\begin{aligned} v_2 &= 3(5) - 2 = 13, \qquad v_3 = 3(13) - 2 = 37, \qquad v_4 = 3(37) - 2 = 109, \quad\ldots \end{aligned}$$ … & so on

$\therefore$ The sequence increases & diverges.

(b)

(ii) $p=1$

Method 1 (GC) (preferred)

Method 2 (algebraic) $$\begin{aligned} v_2 &= 3v_1-2=3(1)-2=1,\\ v_3 &= 3v_2-2=3(1)-2=1,\\ v_4 &= 3v_3-2=3(1)-2=1. \end{aligned}$$ … & so on

$\therefore$ It is a constant sequence which converges to 1.