## H1011 Quizzes

Quiz 10: Summing series
Question 1 Questions
Given the series ${\sum }_{k=1}^{n}\frac{1}{2k\left(2k+1\right)}$ which of the statements below best describes it ?
 a) The series is arithmetic b) The series is geometric c) The series is telescopic d) None of the above

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$\begin{array}{llll}\hfill \sum _{k=1}^{n}\frac{1}{2k\left(2k+1\right)}& =\frac{1}{2k}-\frac{1}{2k+1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\phantom{\rule{1em}{0ex}}\dots -\frac{1}{2n+1},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ so the series looks telescopic but the middle terms don’t cancel out.
Given the series ${\sum }_{k=1}^{n}3+2\left(k-1\right)$, which of the statements below best describes it ?
 a) The series is arithmetic b) The series is geometric c) The series is telescopic d) None of the above

Choice (a) is correct!
${\sum }_{k=1}^{n}3+2\left(k-1\right)$ is arithmetic since it is of the form ${\sum }_{k=1}^{n}a+\left(k-1\right)d$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Given the series ${\sum }_{k=1}^{n}\frac{3}{{2}^{k-1}}$, which of the statements below best describes it ?
 a) The series is arithmetic b) The series is geometric c) The series is telescopic d) None of the above

Choice (a) is incorrect
Choice (b) is correct!
${\sum }_{k=1}^{n}\frac{3}{{2}^{k-1}}={\sum }_{k=1}^{n}3{\left(\frac{1}{2}\right)}^{k-1}$ which is of the form ${\sum }_{k=1}^{n}a{r}^{n-1}$.
Choice (c) is incorrect
Choice (d) is incorrect
Find the sum of the series ${\sum }_{k=0}^{9}\frac{3}{{2}^{k}}$.
 a) $6\left(1-{\left(\frac{1}{2}\right)}^{10}\right)$ b) $6\left(1-{\left(\frac{1}{2}\right)}^{9}\right)$ c) $\frac{3}{2}\left(1-{2}^{10}\right)$ d) None of the above

Choice (a) is correct!
${\sum }_{k=0}^{9}\frac{3}{{2}^{k}}={\sum }_{k=1}^{10}\frac{3}{{2}^{k-1}}$
$=3\left(\frac{{\left(\frac{1}{2}\right)}^{10}-1}{\frac{1}{2}-1}\right)=3\left(\frac{1-{\left(\frac{1}{2}\right)}^{10}}{\frac{1}{2}}\right)=6\left(1-{\left(\frac{1}{2}\right)}^{10}\right)$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the sum of the series ${\sum }_{k=0}^{9}3+2k$
 a) 99 b) 240 c) 198 d) 120

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
${\sum }_{k=0}^{9}3+2k={\sum }_{k=1}^{10}3+2\left(k-1\right)$
$=\frac{10\left(6+9×2\right)}{2}=120.$
Find the sum of the series ${\sum }_{k=1}^{10}\frac{1}{k\left(k+1\right)}$.
 a) $\frac{9}{10}$ b) $1-{\left(\frac{1}{11}\right)}^{10}$ c) $\frac{10}{11}$ d) $1-{\left(\frac{1}{10}\right)}^{10}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
${\sum }_{k=1}^{10}\frac{1}{k\left(k+1\right)}={\sum }_{k=1}^{10}\frac{1}{k}-\frac{1}{k+1}$
$=1-\frac{1}{11}=\frac{10}{11}$.
Choice (d) is incorrect
Write the following in sigma notation
 $2.3.4+3.4.5+4.5.6+\cdots +k\left(k+1\right)\left(k+2\right).$
 a) ${\sum }_{i=1}^{k}\left(i+1\right)\left(i+2\right)\left(i+3\right)$ b) ${\sum }_{i=2}^{k}i\left(i+1\right)\left(i+2\right)$ c) ${\sum }_{i=1}^{k}i\left(i+1\right)\left(i+2\right)$ d) ${\sum }_{i=0}^{k-1}\left(i+1\right)\left(i+2\right)\left(i+3\right)$

Choice (a) is incorrect
Choice (b) is correct!
$\begin{array}{c}\sum _{i=2}^{k}i\left(i+1\right)\left(i+2\right)=2.3.4+3.4.5+\dots +k\left(k+1\right)\left(k+2\right).\end{array}$
Choice (c) is incorrect
Choice (d) is incorrect
Find the sum of the series ${\sum }_{k=0}^{5}\left(2+\frac{2}{{2}^{k}}\right)$.
 a) $\frac{191}{16}$ b) $\frac{223}{16}$ c) $\frac{255}{16}$ d) $\frac{479}{32}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
$\begin{array}{rcll}\sum _{k=0}^{5}\left(2+\frac{2}{{2}^{k}}\right)& =& \sum _{k=0}^{5}2+\sum _{k=0}^{5}\frac{1}{{2}^{k-1}}& \text{}\\ & =& 12+\frac{1}{{2}^{-1}}+\sum _{k=1}^{5}\frac{1}{{2}^{k-1}}& \text{}\\ & =& 14+\frac{31}{16}=\frac{255}{16}.& \text{}\end{array}$
Choice (d) is incorrect
$10000 is deposited in a bank account which pays 6% p.a. compounded every 6 months. The bank also charges$50 in fees every 6 months after calculating the interest. How much is in the account after 3 years ?
 a) 11667.1 b) 11675.1 c) 11731.3 d) 11617.1

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
There are 6 periods of 6 months in 3 years earning 3%. At the end of the period we have $\begin{array}{rcll}{A}_{6}& =& 10000×{\left(1.03\right)}^{6}-50\sum _{i=1}^{6}{\left(1.03\right)}^{i-1}& \text{}\\ & =& 10000×{\left(1.03\right)}^{6}-50\left(\frac{{\left(1.03\right)}^{6}-1}{0.03}\right)& \text{}\\ & =& 11617.10& \text{}\end{array}$
Find the formula for the monthly repayments, $R$, on a 5 year \$10000 car loan at a fixed rate of 15% p.a. compounded monthly.
 a) $R=\frac{10000×0.0125}{1-{\left(1.0125\right)}^{-60}}$ b) $R=\frac{10000×0.0125}{{\left(1.15\right)}^{60}-1}$ c) $R=\frac{10000×0.25}{1-{\left(1.25\right)}^{-60}}$ d) $R=\frac{10000×0.125}{1-{\left(1.125\right)}^{-60}}$

Choice (a) is correct!
The amount owing after $n$ months is
 $10000{\left(1.0125\right)}^{n}-R\left(\frac{1.012{5}^{n}-1}{0.125}\right).$
After 60 months there is nothing owing so
 $R\left(\frac{1.012{5}^{n}-1}{0.125}\right)=10000{\left(1.0125\right)}^{n}.$
After rearranging we get
 $R=\frac{10000×0.0125}{1-{\left(1.0125\right)}^{-60}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect