Number & Algebra

Arithmetic Sequences and Series

  • An arithmetic sequence is a sequence in which the difference between consecutive terms is constant
  • This constant difference is known as the common difference, $d$
  • Example: $d = 3$ for the sequence $1, 4, 7, 10, \ldots$

  • When an arithmetic sequence has a negative common difference, the sequence decreases.
  • Example: $d = -2$ for $10, 8, 6, 4, 2, 0, \ldots$

Finding a Term in an Arithmetic Sequence:


  • The nth term formula for an arithmetic sequence is $u_n=u_1+(n-1)d$
  • Where:
    • $u_1$ is the first term
    • $d$ is the common difference

  • Example: for the arithmetic sequence $4, 9, 14, 19, \ldots$ find the 9th term
    • Solution: $u_9=4+(9-1)5=44$

Sum of an Arithmetic Sequence:


  • An arithmetic series is the sum of the terms in an arithmetic sequence
  • The formulas to find the sum of the first $n$ terms ($S_n$) of an arithmetic sequence are:
    • $S_n = \dfrac{n}{2} (2u_1 + (n-1)\cdot d)$
    • $S_n = \dfrac{n}{2} (u_1 + u_n)$
  • Where:
    • $u_1$ is the first term
    • $d$ is the common difference
    • $u_n$ is the last term

  • Example: For the arithmetic sequence $4, 9, 14, 19, \ldots$ find the sum of the first 9 terms:
    • Solution: $S_9 = \dfrac{9}{2}(4 + 44) = 216$

Applications:


  • One of the most common real-world applications of arithmetic sequences is simple interest, where the same percentage of the initial investment is added each year (as opposed to compound interest).
  • Simple Interest Formula: $\text{Interest} = Prt$, where $P$ is the principal amount, $r$ is the annual interest rate (as a decimal), and $t$ is the number of years.

  • Example: Find the value of a $1200 investment after 6 years with a 4% simple interest rate.
    • Solution: $u_1 = 1200$, $d = 0.04 \cdot 1200 = 48$
    • The problem requires us to find $u_7$, hence $u_7=1200+(6)48=\$1488$

Geometric Sequences and Series

  • A geometric sequence is a sequence in which the ratio between consecutive terms is always equal. In other words, you always multiply by the same constant to move from one term to the next.
  • This constant ratio is known as the common ratio, $r$.
  • Example: For the sequence $1, 3, 9, 27, \ldots$, $r = 3$.

  • When a geometric sequence has a common ratio between 0 and 1, the sequence decreases.
  • Example: $20, 10, 5, 2.5, \ldots$, $r = \dfrac{1}{2}$

  • When a geometric sequence has a negative common ratio, the terms alternate in sign, becoming positive and negative.
  • Example: $96, -48, 24, -12, 6, \ldots$, $r = -\dfrac{1}{2}$

Finding a Term in a Geometric Sequence:


  • The $n$th term formula for a geometric sequence is $u_n = u_1 \cdot r^{n-1}$
  • Where:
    • $u_1$ is the first term
    • $r$ is the common ratio

  • Example: The first 3 terms of a geometric sequence are $0.3$, $1.5$, and $7.5$. Find $u_6$.
    • Solution: $u_6 = 0.3 \cdot 5^5 = 937.5$

Sum of a Geometric Sequence


  • A geometric series is the sum of the terms in a geometric sequence
  • The formulas to find the sum of the first $n$ terms ($S_n$) of a geometric sequence are:
    • $S_n = \dfrac{u_1\left(r^n - 1\right)}{r - 1} \quad$ for $r \neq 1$
    • $S_n = \dfrac{u_1\left(1 - r^n\right)}{1 - r} \quad$ for $r \neq 1$
  • Where:
    • $u_1$ is the first term
    • $r$ is the common ratio
    • $u_n$ is the last term

  • Example: The first 3 terms of a geometric sequence are $0.3$, $1.5$, and $7.5$. Find the sum of the first 6 terms.
    • Solution: $S_6 = \dfrac{0.3\left(5^6 - 1\right)}{5 - 1} = 1,\!171.8$

Diverging Series:


A geometric series is said to diverge when the sum of its terms grows without bound — either toward positive or negative infinity — as more terms are added. This occurs when the absolute value of the common ratio is greater than 1.



Converging Series:


A geometric series converges when the sum of its infinite terms approaches a finite value. This happens only when the absolute value of the common ratio is less than 1.


  • When a geometric sequence is converging (or $r$ is less than 1), we are able to find the sum of infinite terms as the values converge to some value.
  • In order to do this, we can use the equation: $S_∞=\dfrac{u_1}{1-r}$
  • Where:
    • $S_∞$ is the sum of the infinite terms
    • $u_1$ is the first term
    • $r$ is the common ratio

  • Example: If $S_1 = 69.5$ and $S_\infty = 440$ for an infinite geometric sequence, find the common ratio $r$.
    • Solution: $440 = \dfrac{69.5}{1 - r}$ $\implies$ $r = 1 - \dfrac{69.5}{440} = 0.842$

Applications:


  • The main application of geometric sequences is compound interest, however they may also be applied to exponential population growth and decay, spread of disease, depreciation in the value of assets and more.

  • Compound Interest is when interest is calculated on both the principal amount of money, and on the interest payable or earned on that principal amount.
  • Compound Interest Formula: $F.V. = P.V. \left(1 + \dfrac{r}{100k}\right)^{kn}$
  • Where:
    • $F.V.$ is the future value
    • $P.V.$ is the present value
    • $r$ is the interest rate per year
    • $k$ is the number of times interest is compounded per year
    • $n$ is the number of years

  • Example 1: Find the value of a $\$5000$ investment if compounded annually with an interest rate of $4.2%$ per annum after $5$ years.
    • Solution: $\text{Final Value} = 5000 \cdot (1.042)^5 = \$6141.98$

  • Example 2: A $\$4500$ investment gives $\$5,533.04$ if compounded quarterly for $4$ years at what % rate per annum?
    • Solution: $5533.04 = 4500 \left(1 + \dfrac{r}{400}\right)^{16}$
    • Hence: $1.2296\ldots = \left(1 + \dfrac{r}{400}\right)^{16}$
    • And: $r = 5.2\%$

The Binomial Theorem

  • In this section, we will be dealing with any kind of expansion in the form: $(a+b)^n$
  • For example:
    • $(a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2$
    • $(a+b)^3 = (a+b)(a+b)(a+b) = a^3 + 3a^2b + 3ab^2 + b^3$

  • As you can see above, there are specific coefficients that multiply with each term. For $(a+b)^2$: these coefficients are 1, 2 and 1. For $(a+b)^3$: these coefficients are 1, 3, 3 and 1. These coefficients will depend on the power ($n$), and which powers of $a$ & $b$ you are considering.

  • Let's consider how we found $3ab^2$: This was all the different ways of choosing two $b$'s from the three brackets. For any situation, if we say $r$ is the power of $b$, then this coefficient will be equal to the number of ways to choose $r$ from the $n$ total objects.

    • In order to do this, we can use "$n$ CHOOSE $r$"
    • This represents the number of ways to choose a subset of $r$ elements from a set of $n$ distinct elements, where the order of selection does not matter.
    • "$n$ CHOOSE $r$" can be notated as ${}_nC_r$ and calculated using the formula: ${}_nC_r = \dfrac{n!}{r!(n-r)!}$, where $n! = 1 \cdot 2 \cdot 3 \cdots (n-1) \cdot n$
    • Example: ${}_7C_5 = \dfrac{7!}{5! \cdot 2!} = \dfrac{5040}{120 \cdot 2} = 21$
    • The ${}_nC_r$ function can be done on your GDC, which is helpful if the numbers start getting quite big. For TI-84: Math >> PRB > 3: nCr >> Type: nCr(n,r) >> ENTER

  • To find a specific term in a binomial expansion, use the general term formula: ${}_nC_r \cdot a^{n-r} \cdot b^r$

Pascal's Triangle:


  • In this seemingly unrelated pattern, where every number is found by adding the two numbers above, each set of values for $n$ is represented by each row. Then, the leftmost value is for $r=0$, and as you move right, $r$ increases.



  • Hence, it can be used to find coefficients in the expansion of any binomial expression:



  • Now that we know how to calculate the ${}_nC_r$ coefficients, we can put together the whole expansion, with descending powers of $a$ and ascending powers of $b$:

    • $$(a+b)^n = {}_nC_0\cdot a^n\cdot b^0 + {}_nC_1\cdot a^{n-1}\cdot b^1 + {}_nC_2\cdot a^{n-2}\cdot b^2 + \ldots + {}_nC_n\cdot a^0\cdot b^n$$


  • On exams, questions often ask you to find one specific term out of the expansion, such as the $x^3$ term or the "constant term".

  • Example 1: Fully expand $(2x+3)^4$:
    • Solution: $(2x+3)^4 = (2x)^4 + {}_4C_1 (2x)^3 3^1 + {}_4C_2 (2x)^2 3^2 + {}_4C_3 (2x)^1 3^3 + 3^4$
    • Hence: $(2x+3)^4 = 16x^4 + 96x^3 + 216x^2 + 216x + 81$

  • Example 2: Find the constant term, with $x^0$, in the expansion of $\left(2x-\frac 4x\right)^6$:
    • Solution: ${}_6C_3\cdot (2x)^3\cdot \left(-\dfrac 4x\right)^3 = 20 \cdot 8x^3 \cdot -\dfrac{64}{x^3} = -10240x^0 = -10240$

Sigma Notation

Sigma notation is a way to represent the summation of any sequence — this means that it can be used for both arithmetic or geometric series. The notation shows you the formula that generates terms of a sequence and the upper and lower limits of the terms that you want to add up in this sequence.


Notation:

$$\Large \sum_{i=1}^{n} x_i$$


Where:


  • $\sum$ is the Greek letter sigma, meaning "sum" in this case
  • $i$ is the index of summation, which starts at $1$ and goes up to $n$
  • $x_i$ is the general term in the sequence that you want to sum
  • $n$ is the upper limit of the summation

Examples:

$$\large \sum_{i=1}^{5} i^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$$

$$\large \sum_{i=1}^{n} i = 1 + 2 + 3 + \ldots + n = S_n \quad (\text{arithmetic series}) = \dfrac n2 (1+n)$$