Summation Notation

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}} \DeclareMathOperator {\proj }{\vec {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

We review the basic of summation notation.

Consider the expression: $\sum _{i=a}^b c_i$ Here each $c_i$ is a element factor index letter lower limit term upper limit in the sum, $i$ is called the element factor index letter lower limit term upper limit of summation, $a$ is called the element factor index letter lower limit term upper limit of summation, and $b$ is called the element factor index letter term upper limit of summation.

Write the terms of this series: $\sum _{i=1}^4 \left (7-2i\right )=\answer {5}+\answer {3}+\answer {1}+\answer {-1}$

Write the terms of this series: $\sum _{i=1}^4 \left (3i^2-5\right )=\answer {-2}+\answer {7}+\answer {22}+\answer {43}$

Express the sum $1^2 + 2^2 + 3^2 + 4^2+\dots +27^2$ in summation notation: $\sum _{k=1}^{\answer {27}} \answer {k^2}$

Express the sum $2 + 2^2 + 2^3 + 2^4+\dots +2^{11}$ in summation notation: $\sum _{n=1}^{\answer {11}} \answer {2^n}$

Express the sum $7 + \frac {7^2}{2} + \frac {7^3}3 + \frac {7^4}{4}+\dots +\frac {7^{43}}{43}$ in summation notation: $\sum _{i=1}^{\answer {43}} \answer {7^i/i}$

Verify that $\left (\sum _{k = 1}^2 k\right )^2 \ne \sum _{k = 1}^2 k^2$ by computing the value of both sides: $\begin{align*} \left(\sum_{k = 1}^2 k\right)^2 &= \answer{9} \\ \sum_{k = 1}^2 k^2 &= \answer{5} \end{align*}$

The following are important algebraic rules of exponents:

(a): $(ab)^n = a^n b^n$
(b): $\left ( \frac {a}{b} \right )^n = \frac {a^n}{b^n}$
(c): $(a^n)^m = a^{nm}$
(d): $a^n a^m = a^{n+m}$
(e): $\frac {a^n}{a^m} = a^{n - m}$
(f): $\sqrt [n]{a^m} = a^{m/n}$ .

In the following problems, you will be asked to simplify expressions involving exponents.

I understand I do not understand

One of the most important results that will be introduced is a shortcut for adding the terms in a special type of series, called a geometric series. The formula that will be derived in class states: $\sum _{k = 0}^n a r^k = \frac {a - ar^{n+1}}{1 - r}$ where $a$ and $r$ are constants.

In order to use this formula, it is necessary to be able to write a given expression in this form by using the laws of exponents.

For example: $2^{3k - 1} = 2^{3k} \cdot 2^{-1} = (2^3)^k \cdot \frac {1}{2} = 8^k \cdot \frac {1}{2} = \frac {1}{2} \cdot 8^k$ (Here, $a = 1/2$ and $r = 8$ .)

The next few questions give practice doing this.

I understand I do not understand

Write the expression $\frac {4^{k+1}}{3^{2k}}$ in the form $a r^k$ by identifying $a$ and $r$ : $\begin{align*} a &= \answer{4} \\ r &= \answer{4/9} \end{align*}$

Write the expression $\frac {3^{2k + 1}}{2 \cdot 2^k}$ in the form $a r^k$ by identifying $a$ and $r$ : $\begin{align*} a &= \answer{3/2}\\ r &= \answer{9/2} \end{align*}$

Select ALL of the statements below that are algebraically correct.

$\frac {4x^2 - 2x}{3x} = \frac {x(4x-2)}{3x} = \frac {4x - 2}{3}$ $\frac {2x}{1+x^2} = \frac {2x}{1} + \frac {2x}{x^2} = 2x + \frac {2}{x}$ $(x - 1)^3 = x^3 - 1$ $\frac {1}{5x} = 5x^{-1}$ $(2x)^3 = 2^3 x^3 = 8x^3$ $2^{3 + 2k} = 2^3 + 2^{2k}$