Sequences Series Part3

Exercise 1

1) Find the following sums

a) $\sum_{r=1}^{n}\left(2r-1\right)$ $n^{2}\,$
- $\begin{align} \sum_{r=1}^{n}\left(2r-1\right) & = 2\sum_{r=1}^{n}r-\sum_{r=1}^{n}1\\ & = 2\left[\frac{1}{2}n\left(n+1\right)\right]-n \\ & =n\left(n+1\right)-n \\ & = n^{2} \end{align}$
- Check :
  - $\begin{array}{l|l|l} n=1 & 1 & 1^{2} \\ n=2 & 1+3 & 2^{2} \\ \end{array}$

b) $\sum_{r=1}^{n}\left(r+1\right)\left(r+2\right)$ $\frac{1}{3}n\left(n^{2}+6n+11\right)$
- $\begin{align} \sum_{r=1}^{n}\left(r+1\right)\left(r+2\right) & = \sum_{r=1}^{n}\left(r^{2}+3r+2\right)\\ & =\sum_{r=1}^{n}r^{2}+3\sum_{r=1}^{n}r+\sum_{r=1}^{n}2\\ & = \frac{1}{6}n\left(n+1\right)\left(2n+1\right) + 3\left[\frac{1}{2}n\left(n+1\right)\right]+2n \\ & = \frac{1}{6}n\left[\left(n+1\right)\left(2n+1\right)+9\left(n+1\right)+12\right]\\ & = \frac{1}{6}n\left(2n^{2}+3n+1+9n+9+12\right)\\ & = \frac{1}{6}n\left(2n^{2}+12n+22\right)\\ & = \frac{1}{3}n\left(n^{2}+6n+11\right) \end{align}$
- Check :
  - $\begin{array}{l|l|l} n=1 & 2\cdot3 & \frac{1}{3}\left(18\right) \\ \end{array}$

c) $\sum_{r=1}^{n}r\left(r^{2}+1\right)$ $\frac{1}{4}n\left(n+1\right)\left(n^{2}+n+2\right)$
- $\begin{align} \sum_{r=1}^{n}r\left(r^{2}+1\right) & = \sum_{r=1}^{n}\left(r^{3}+r\right)\\ & =\sum_{r=1}^{n}r^{3}+\sum_{r=1}^{n}r\\ & = \frac{1}{4}n^{2}\left(n+1\right)^{2}+\frac{1}{2}n\left(n+1\right) \\ & = \frac{1}{4}n\left(n+1\right)\left[n\left(n+1\right)+2\right]\\ & = \frac{1}{4}n\left(n+1\right)\left(n^{2}+n+2\right)\\ \end{align}$
- Check :
  - $\begin{array}{l|l|l} n=1 & 1\left(1^{1}+1\right) & \frac{1}{4}\left(1\right)\left(2\right)\left(4\right) \\ \end{array}$

d) $\sum_{r=1}^{20}\left(r^{3}+r^{2}+r+1\right)$ $47200\,$
- $\begin{align} \sum_{r=1}^{20}\left(r^{3}+r^{2}+r+1\right) & = \sum_{r=1}^{20}r^{3}+\sum_{r=1}^{20}r^{2}+\sum_{r=1}^{20}r+\sum_{r=1}^{20}1\\ & = \frac{1}{4}\left(20\right)^{2}\left(21\right)^{2}+\frac{1}{6}\left(20\right)\left(21\right)\left(41\right)+\frac{1}{2}\left(20\right)\left(21\right)+20 \\ & = 47200 \end{align}$

e) $\sum_{r=1}^{12}4^{r}$ $22369620\,$
- $\begin{align} \sum_{r=1}^{12}4^{r} & = \underbrace{4+4^{2}+4^{3}+\ldots +4^{12}}_{\mbox{GP,}a=4, r=4, n=12 }\\ & =\frac{4\left(4^{12}-1\right)}{4-1}\\ & =22369620 \\ \end{align}$

f) $\sum_{r=1}^{8}3^{2r}$ $48427560\,$
- $\begin{align} \sum_{r=1}^{8}3^{2r} & = \underbrace{3^{2}+3^{4}+3^{6}+\ldots +3^{16}}_{\mbox{GP,}a=9, r=9, n=8 }\\ & =\frac{9\left(9^{8}-1\right)}{9-1}\\ & =48427560 \\ \end{align}$

g) $\sum_{r=1}^{n}3\left(2^{r-1}\right)$ $3\left(2^{n}-1\right)\,$
- $\begin{align} \sum_{r=1}^{n}3\left(2^{r-1}\right) & = 3\sum_{r=1}^{n}2^{r-1}\\ & = 3\underbrace{\left(1+2+4+\ldots +2^{n-1}\right)}_{\mbox{GP,}a=1, r=2, n \mbox{ terms}}\\ & =3\left[\frac{\left(2^{n}-1\right)}{2-1}\right]\\ & =3\left(2^{n}-1\right) \\ \end{align}$
- Check :
  - $\begin{array}{l|l|l} n=1 & 3 & 3 \\ n=2 & 3+6 & 3\left(3\right) \\ \end{array}$

h) $\sum_{r=1}^{n}\left[\left(\frac{1}{2}\right)^{3r}+3\right]$ $\frac{1}{7}\left(1-\frac{1}{8^{n}}\right)+3n\,$
- $\begin{align} \sum_{r=1}^{n}\left[\left(\frac{1}{2}\right)^{3r}+3\right] & = \sum_{r=1}^{n}\left(\frac{1}{2}\right)^{3r}+\sum_{r=1}^{n}3 \\ & =\underbrace{\left(\frac{1}{2^{3}}+\frac{1}{2^{6}}+\frac{1}{2^{9}}+\ldots +\frac{1}{2^{3n}}\right)}_{\mbox{GP,}a=\frac{1}{8}, r=\frac{1}{8}, n \mbox{ terms} }+3n\\ & =\frac{\frac{1}{8}\left[1-\left(\frac{1}{8}\right)^{n}\right]}{1-\frac{1}{8}}+3n\\ & =\frac{1}{7}\left(1-\frac{1}{8^{n}}\right)+3n\ \\ \end{align}$
- Check :
  - $\begin{array}{l|l|l} n=1 & \frac{1}{8}+3 & \frac{1}{7}\left(\frac{7}{8}\right)+3 \\ \\ n=2 & \frac{1}{8}+3+\frac{1}{64}+3 & \frac{1}{7}\left(\frac{63}{64}\right)+6 \\ \end{array}$

i) $\sum_{r=1}^{15}\left(2^{r}+3^{r-1}\right)$ $7239987\,$
- $\begin{align} \sum_{r=1}^{15}\left(2^{r}+3^{r-1}\right) & = \sum_{r=1}^{15}2^{r}+\sum_{r=1}^{15}3^{r-1}\\ & = \underbrace{\left(2+2^{2}+2^{3}+\ldots +2^{15}\right)}_{\mbox{GP,}a=2, r=2, n=15 }+\underbrace{\left(1+3+3^{2}+\ldots +3^{14}\right)}_{\mbox{GP,}a=1, r=3, n=15}\\ & =\frac{2\left(2^{15}-1\right)}{2-1}+\frac{1\left(3^{15}-1\right)}{3-1}\\ & = 65534 + 7174453 \\ & = 7239987 \end{align}$

2) Evaluate the following series

a) $1 \cdot 2+ 2\cdot 3+ 3 \cdot 4+ \ldots + 100 \cdot 101$ $343400\,$
$\begin{align} & u_{r} =r\left(r+1\right), n=100 \\ & 1 \cdot 2+ 2\cdot 3+ 3 \cdot 4+ \ldots + 100 \cdot 101 \\ & = \sum_{r=1}^{100} r\left(r+1\right) \\ & = \sum_{r=1}^{100} \left(r^{2}+r\right) \\ & = \sum_{r=1}^{100}r^{2}+\sum_{r=1}^{100}r \\ & =\frac{1}{6}\left(100\right)\left(101\right)\left(201\right)+\frac{1}{2}\left(100\right)\left(101\right)\\ & = 338350 + 5050 \\ & =343400\\ \end{align}$

b) $2^{2} \cdot 3+ 3^{2}\cdot 5+ 4^{2} \cdot 7+ \ldots + 50^{2} \cdot 99$ $3208324\,$