A quadratic equation is a second-order polynomial equation in a single variable $x$.

$ax^2+bx+c=0$ -------- (1)

with $a \neq 0$

Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has two solutions. These solutions may be both real or complex.

The roots x can be found by completing the square,

$x^2+\dfrac{b}{a}x=-\dfrac{c}{a}$ -------- (2)

$\left(x+\dfrac{b}{2a}\right)^2=-\dfrac{c}{a}+\dfrac{b}{4a}^2$ $=\dfrac{b^2-4ac}{4a^2}$ -------- (3)

$x+\dfrac{b}{2a}= \pm \dfrac{\sqrt{b^2-4ac}}{2a}$ -------- (4)

Solving for $x$ then gives

$x=\dfrac{-b \pm \sqrt {b^2-4ac}}{2a}$ -------- (5)

This equation is known as the quadratic formula. The letters $a$, $b$ and $c$ are coefficients (we know those values). They can have any value, except that can't be 0. The letter $x$ is the variable or unknown (we don't know it yet). Here are some more examples:

Table below can be scrolled horizontally

 $2x^2+5x+3=0$ In this one $a=2$,$b=5$ and$c=3$ $x^2-3x=0$ This one is a little more tricky: Where is a? In fact $a=1$, as we don't usually write "$1x^2$" And where is c? Well, $c=0$, so is not shown. $4x-3=0$ Oops! This one is not a quadratic equation, because it is missing $x^2$ (in other words $a=0$, and that means it can't be quadratic)

The graph it makes and the solutions (called "roots") are shown in the diagram below:

Â

The "solutions" to the Quadratic Equation are where it is equal to zero. There are usually 2 solutions (as shown in the graph above). They are called "roots", or sometimes "zeros"

The $\pm$ means there are TWO answers:

$x=\dfrac{-b+ \sqrt {b^2-4ac}}{2a}$

$x=\dfrac{-b- \sqrt {b^2-4ac}}{2a}$

But sometimes you don't get two real answers, and the "Discriminant" shows why ...

### 1.1 Discriminant

Do you see $b^2-4ac$ in the formula above? It is called the Discriminant, because it can "discriminate" between the possible types of answer:

When $b^2-4ac$ is positive, you get two Real solutions

When it is zero you get just ONE real solution (both answers are the same)

When it is negative you get two Complex solutions

#### Example:

Solve $5x^2+6x+1=0$

#### Solution:

Coefficients are $a=5$, $b=6$, $c=1$

Quadratic Formula: $x=\dfrac{-b \pm \sqrt {b^2-4ac}}{2a}$

Put in $a$, $b$ and $c$: $x=\dfrac{-6 \pm \sqrt {6^4-4 \times 5 \times 1}}{2 \times 5}$

$x = \dfrac{-6 \pm \sqrt {(36-20)} }{10}$

$x = \dfrac{-6 \pm \sqrt {(16)} }{10}$

$x = \dfrac{-6 \pm 4 }{10}$

$x = -0.2$ or $-1$

$x = -0.2$ or $x = -1$

And you can see them on the graph

\begin{align*} \text{Check -0.2:}\\ &=5 \times (-0.2)^2 + 6 \times (-0.2) + 1\\ &= 5 \times (0.04) + 6 \times (-0.2) + 1 \\ &= 0.2 -1.2 + 1\\ &= 0\\ \text{Check -1.0:}\\ &=5 \times (-1)^2 + 6 \times (-1) + 1\\ &= 5 \times (1) + 6 \times (-1) + 1\\ &= 5 - 6 + 1\\ &= 0 \end{align*}

### 1.2 Complex Solutions?

When the Discriminant (the value $b^2-4ac$) is negative you get Complex solutions.

What does that mean?

#### Example:

Solve $5x^2+6x+1=0$

#### Solution:

Coefficients are $a=5$, $b=6$,$c=1$

Note that The Discriminant is negative:

$b^2 - 4ac = 2^2 - 4 \times 5 \times 1 = -16$

$x = \dfrac{ -2 \pm \sqrt (-16) }{10}$

The square root of $-16$ is $4i$

($i$ is $\sqrt{-1}$, read Imaginary Numbers to find out more), So:

$x = \dfrac{ -2 \pm 4i }{10}$

$x = -0.2 \pm 0.4i$

The graph does not cross the x-axis. That is why we ended up with complex numbers.

### 1.3 Remainder Theorem

To identify whether a given expression is a factor of another expression, we can take help of Remainder Theorem.

According to the remainder theorem, when any expression$f(x)$ is divided by $(x-a)$, the remainder is $f(a)$ ($a$ is any constant in this example).

Thus when the expression, $x^3+x^2+4$ is divided by $x+1$, the remainder is:

$(-1)^3+(-1)^2+4=4$

### 1.4 Factor Theorem

We can also use the remainder theorem or more specifically the Factor theorem to identify if an expression is a factor of another expression.

As already seen, a expression is is a said to be a factor of another expression only when the remainder is 0 when the latter is divided by the former.

Thus, $(x-a)$ is a factor of $f(x)$ if and only if $f(a)=0$.This theorem also helps us in factorizing higher degree equations. Consider the equation:

$f(x)=x^3+6x^2-19x-24=0$

By hit and trial we see that:

$f(-1)=(-1)^3+6(-1)^2-19(-1)-24=0$

Thus, we can deduce that $(x+1)$ is a factor of $f(x)$. i.e.

$x^3+6x^2-19x-24=(x+1) \times g(x)$

Where $g(x)$ is another algebraic expression in variable $x$.

## 2. Functions

### 2.1 Defining the Graph of a Function

The graph of a function $f$ is the set of all points in the plane of the form $(x, f(x))$. We could also define the graph off to be the graph of the equation $y = f(x)$. So the graph of a function is a special case of the graph of an equation.

#### Example:

Let $f(x)=x^2-3$

Recall that when we introduced graphs of equations we noted that if we can solve the equation for $y$, then it is easy to find points that are on the graph. We simply choose a number for $x$, then compute the corresponding value of $y$. Graphs of functions are graphs of equations that have been solved for $y$!

The graph of $f(x)$ in this example is the graph of $y = x^2 - 3$. It is easy to generate points on the graph. Choose a value for the first coordinate, then evaluate f at that number to find the second coordinate. The following table shows several values for $x$ and the function f evaluated at those numbers.

Table below can be scrolled horizontally

 $x$ -2 -1 0 1 2 $f(x)$ 1 -2 -3 -2 1

Each column of numbers in the table holds the coordinates of a point on the graph.

#### Example:

Let $f$ be the piecewise-defined function

$$f(x) = \begin{cases} \nonumber 5-x^2 & x\leq 2\\ x-1 & x > 2 \end{cases}$$

The graph of $f$ is shown below.

### 2.2 Some of the Important Functions

Table below can be scrolled horizontally

Function Definition Example
Linear

The highest power over the $x$ variable is 1

$y = 2x + 1$

The highest power over the $x$ variable(s) is 2

$y = -2x^2-7x+2$

Cubic

The highest power over the $x$ variable(s) is 3

$y= x^3 + 2x^2+x$

Square Root

The $x$ variable is square rooted

$y=- \sqrt{x-1}$

Absolute Value

The $x$ variable is within absolute value signs

$y=-|x+1|$

Rational

The $x$ variable is in the denominator $(x \neq 0 )$

$y=\dfrac{3}{x}$

Logarithmic

The log of the $x$ variable is taken

$\ln(x)$

Graph of $y=x$ (Linear)

Graph of $y=x^2$ (Quadratic)

Graph of $y=x^3$ (Cubic)

Graph of $y= \sqrt x$ (Square root)

Graph of $y=|x|$ (Absolute value)

Graph of $y=\dfrac{1}{x}$ (Rational)

Graph of $y=\ln(x)$ (Logarithmic)

### 2.3 Even and Odd Functions

A function f is even if its graph is symmetric with respect to the y-axis. This criterion can be stated algebraically as follows: f is even if $f(-x) = f(x)$ for all $x$ in the domain of $f$.

#### Example:

If you evaluate $f$ at 3 and at -3, then you will get the same value if $f$ is even.

A function $f$ is odd if its graph is symmetric with respect to the origin. This criterion can be stated algebraically as follows: $f$ is odd if $f(-x) = -f(x)$ for all $x$ in the domain of $f$.

#### Example:

If you evaluate $f$ at 3, you get the negative of $f(-3)$ when $f$ is odd.

## 3. Inequalities

An inequality is the result of replacing the "$=$" sign in an equation with $, \leq ,\text{ or }\geq$.

For example, $3x - 2 0$, are referred to as polynomial inequalities, or quadratic inequalities if the degree is exactly 2.

Inequalities involving rational expressions are called rational inequalities. Some often used inequalities also involve absolute value expressions.

### 3.1 Solving Inequalities

In a nutshell, solving inequalities is about one thing: sign changes. Find all the points at which there are sign changes - we call these points' critical values.

Then determine which, if any, of the intervals bounded by these critical values result in a solution.

The solution to the inequality will consist of the set of all points contained by the solution intervals.

### 3.2 Method to Solve Linear, Polynomial or Absolute Value Inequalities

1. Move all terms to one side of the inequality sign by applying the Addition, Subtraction, Multiplication, and Division Properties of Inequalities. You should have only zero on one side of the inequality sign.

2. Solve the associated equation using an appropriate method. This solution or solutions will make up the set of critical values. At these values, sign changes occur in the inequality.

3. Plot the critical values on a number line. Use closed circles $\bullet$ for $\leq$ and $\geq$ inequalities, and use open circles $\circ$ for $$inequalities. 4. Test each interval defined by the critical values. If an interval satisfies the inequality, then it is part of the solution. If it does not satisfy the inequality, then it is not part of the solution. #### Example: Solve:  3x + 5(x + 1) \leq 4x - 1 and graph the solution. #### Solution: 3x + 5(x + 1) \leq 4x - 1 3x + 5x + 5 \leq 4x - 1  8x + 5 \leq 4x - 1  4x + 6 \leq 0 Now, solve 4x+6 = 0 4x = -6  x = -3/2 Test arbitrary values of each interval. You may test the value in the original Inequality or its simplified form. In Interval 1, let x=1 in 4x + 6 \leq 0. 4 \times (1) + 6 \leq 0 is FALSE. In Interval 2, let x = -2 in 4x + 6 \leq 0. 4 \times (-2) + 6 \leq 0 is TRUE. So Interval 2 is the solution. The solution is x \leq - 3/2, or in interval notation, (- \infty , - 3/2]. #### Example: Solve 3x^3+5x^2>4x^2 and graph the solution #### Solution: 3x^3+5x^2>4x^2 3x^3+ x^2> 0 Now, solve 3x^3 + x^2 = 0 x^2(3x + 1) = 0 x^2=0 or  3x + 1 = 0 Solve x^2=0 x = 0 Solve 3x + 1 = 0 3x = -1 x = -1/3 Plot the critical values, x = 0 and  x =-1/3. Use open circles this time! Test arbitrary values of each interval. You may test the value in the original inequality or its simplified form. In Interval 1, letx=1 in x^2(3x+1)>0 (3 \times 1 + 1) > 0 is TRUE, so Interval 1 is part of the solution. In Interval 2, let x=-1/4 in x^2(3x+1)>0 (3 \times \left(-\dfrac{1}{4}\right) + 1) > 0 is TRUE, since simplified, we get \left(\dfrac{1}{16}\right)\left(\dfrac{1}{4}\right) > 0, So interval 2 is the part of the solution. In interval 3, let x=-1 in x^2(3x+1)>0 (3 \times (-1) + 1) > 0 is FALSE, since simplified, we get (1) \times (-2) > 0, So Interval 3 is NOT part of the solution. We shade Interval 1 and Interval 2, but do not include the endpoints. The solution is x>0 or -1/3 The interval notation of this solution is (0 , \infty ) \cup (-1/3, 0) #### Example: Solve |2x + 5|> 1 and graph the solution. #### Solution: Given, |2x + 5|> 1 Solve the related equation |2x + 5| = 1 To solve absolute value equations, you must solve two cases: 2x + 5 = 1 and -(2x + 5) = 1 Solve 2x + 5 = 1 2x = -4 x = -2 Solve -(2x + 5) = 1 -2x - 5 = 1 -2x = 6 x = -3 Plot the critical values, x = -2 and x =-3. Test arbitrary values of each interval. You may test the value in the original Inequality or its simplified form. In Interval 1, let x=0 in|2x + 5|> 1. |2 \times 0 + 5|> 1 is TRUE, so Interval 1 is part of the solution. In Interval 2, let x = -2.5 in |2x + 5|> 1 |2 \times (-2.5) + 5> 1 is FALSE, so Interval 2 is NOT part of the solution. In Interval 3, let x=- 4 in|2x + 5|> 1. |2 \times (-4) + 5|> 1is TRUE, so Interval 3 is part of the solution. We shade Interval 1 and Interval 3, but do not include the endpoints. ### 3.3 Method to Solve Rational Inequalities 1. Move all terms to one side of the inequality sign by applying the Addition, Subtraction, Multiplication, and Division Properties of Inequalities. You should have only zero on one side of the inequality sign. 2. Solve the associated equation using an appropriate method. This solution or solutions will make up the set of critical values. At these values, sign changes occur in the inequality. 3. Find all values that result in Division by Zero. These are also critical values for rational inequalities. 4. Plot the critical values on a number line. Use closed circles \bullet for \leq and \geq unless the value results in division by zero - always use open circles for values resulting in division by zero since this value cannot be part of the solution! Always use open circles \circ for$$ inequalities.

5. Test each interval defined by the critical values. If an interval satisfies the inequality, then it is part of the solution. If it does not satisfy the inequality, then it is not part of the solution.

## 4. Progressions

Succession of numbers of which one number is designated as the first, other as the second, another as the third and so on gives rise to what is called a sequence.

Sequences have wide applications. In this lesson we shall discuss particular types of sequences called arithmetic sequence, geometric sequence and also find arithmetic mean (A.M), geometric mean (G.M) between two given numbers.It is a set of numbers which are written in some particular order. For example, take the numbers

$1, 3, 5, 7, 9, ... .$

Here, we seem to have a rule. We have a sequence of odd numbers. To put this in another way, we start with the number 1, which is an odd number, and then each successive number is obtained by adding 2 to give the next odd number.

Here is another sequence:

$1, 4, 9, 16, 25, ... .$

This is the sequence of square numbers. And this sequence,

$1, -1, 1, -1, 1, -1,....$

is a sequence of numbers alternating between 1 and -1. In each case, the dots written at the end indicate that we must consider the sequence as an infinite sequence, so that it goes on forever.

On the other hand, we can also have finite sequences. The numbers

$1, 3, 5, 9$

form a finite sequence containing just four numbers. The numbers

$1, 4, 9, 16$

also form a finite sequence. And so do these, the numbers

$1, 2, 3, 4, 5, 6, ... , n$

### 4.1 Series

A series is something we obtain from a sequence by adding all the terms together.

#### Example:

Suppose we have the sequence

$u_1, u_2, u_3, ... ,u_n$

The series we obtain from this is

$u_1 + u_2 + u_3 + ... + u_n$

and we write S_n for the sum of these n terms. So although the ideas of a 'sequence' and a 'series' are related, there is an important distinction between them.

#### Example:

Let us consider the sequence of numbers

$1, 2, 3, 4, 5, 6, ....n$

Then $S_1 = 1$, as it is the sum of just the first term on its own. The sum of the first two terms is:

$S_2 = 1 + 2 = 3$.

Continuing, we get:

$S_3 = 1 + 2 + 3 = 6$,

$S_4 = 1 + 2 + 3 + 4 = 10$

### 4.2 Arithmetic Progressions

Consider these two common sequences

$1, 3, 5, 7, ...$and $0, 10, 20, 30, 40, ...$

It is easy to see how these sequences are formed. They each start with a particular first term, and then to get successive terms we just add a fixed value to the previous term. In the first sequence we add 2 to get the next term, and in the second sequence we add 10. So the difference between consecutive terms in each sequence is a constant.

We could also subtract a constant instead, because that is just the same as adding a negative constant. For example, in the sequence

$8, 5, 2, -1, -4, ...$

the difference between consecutive terms is -3. Any sequence with this property is called an arithmetic progression, or AP for short.

We can use algebraic notation to represent an arithmetic progression. We shall let a stand for the first term of the sequence, and let d stand for the common difference between successive terms.

For example, our first sequence could be written as

$1, 3, 5, 7, 9, ...$

$1, (1 + 2), (1 + 2 \times 2), (1 + 3 \times 2), (1 + 4 \times 2), ... ,$

and this can be written as

$a, a + d, a + 2d, a + 3d, a + 4d, ...$

where, $a = 1$ is the first term, and $d = 2$ is the common difference. If we wanted to write down the $n^{th}$ term, we would have

$a + (n - 1)d$,

We also sometimeswrite $l$ for the last term of a finite sequence, and so in this case we would have

$l = a + (n - 1)d$ .

### 4.3 The Sum of an Arithmetic Series

Sometimes we want to add the terms of a sequence. What would we get if we wanted to add the first n terms of an arithmetic progression? We would get

$S_n = a + (a + d) + (a + 2d) + ... + (l - 2d) + (l - d) + l$.

Now this is now a series, as we have added together the n terms of a sequence. This is an arithmetic series, and we can find its sum by using a trick. Let us write the series down again, but this time we shall write it down with the terms in reverse order. We get

$S_n = l + (l - d) + (l - 2d) + ... + (a + 2d) + (a + d) + a$.

We are now going to add these two series together. On the left-hand side, we just get $2S_n$. But on the right-hand side, we are going to add the terms in the two series so that each term in the first series will be added to the term vertically below it in the second series. We get

$2S_n = (a + l) + (a + l) + (a + l) + ... + (a + l) + (a + l) + (a + l)$

and on the right-hand side there are n copies of $(a + l)$ so we get

$2S_n = n(a + l)$.

But of course we want $S_n$ rather than $2S_n$, and so we divide by 2 to get

$S_n =\dfrac{1}{2} n(a + l)$

We have found the sum of an arithmetic progression in terms of its first and last terms, $a$ and $l$, and the number of terms $n$.

We can also find an expression for the sum in terms of the $a$, $n$ and the common difference $d$. To do this, we just substitute our formula for $l$ into our formula for $S_n$. From:

$l = a + (n - 1)d$ , $S_n =\dfrac{1}{2} n(a + l)$

we obtain

$S_n =\dfrac{1}{2} n(a +a+(n-1) d)$

$S_n =\dfrac{1}{2} n(2a+(n-1) d)$

#### Example:

Find the sum of the first 50 terms of the sequence

$1, 3, 5, 7, 9, ... .$

#### Solution:

This is an arithmetic progression, and we can write down

$a = 1$ ,$d = 2$ , $n = 50$ .

We now use the formula, so that

$S_n =\dfrac{1}{2} n(2a+(n-1) l)$

$S_{50} =\dfrac{1}{2} \times 50 \times (2 \times 1 + (50 - 1) \times 2)$

$= 25 \times (2 + 49 \times 2)$

$= 25 \times (2 + 98)$

$= 2500$ .

#### Example:

Find the sum of the series

$1 + 3 \times 5 + 6 + 8 \times 5 + ... + 101$

#### Solution:

This is an arithmetic series, because the difference between the terms is a constant value, 2.5

We also know that the first term is 1, and the last term is 101. But we do not know how many terms are in the series. So we will need to use the formula for the last term of an arithmetic progression,

$l = a + (n - 1)d$

to give us

$101 = 1 + (n - 1) \times 2.5$

Now this is just an equation for n, the number of terms in the series, and we can solve it. If we subtract 1 from each side we get

$100 = (n - 1) \times 2.5$

and then dividing both sides by 2.5 gives us $40 = n - 1$so that $n = 41$. Now we can use the formula or the sum of an arithmetic progression, in the version using l, to give us

$S_n= \dfrac{1}{2} n(a + l)$

$S_{41}=\dfrac{1}{2} \times 41 \times (1+101)$

$=\dfrac{1}{2} \times 41 \times 101$

$= 2091$

#### Example:

An arithmetic progression has 3 as its first term. Also, the sum of the first 8 terms is twice thesum of the first 5 terms. Find the common difference.

#### Solution:

We are given that $a = 3$. We are also given some information about the sums $S_8$ and $S_5$, and we want to find the common difference. So we shall use the formula

$S_n=\dfrac{n}{2}(2a + (n - 1)d)$

for the sum of the first $n$ terms. This tells us that

$S_8 =\dfrac{1}{2} \times 8 \times (6 + 7d)$

$S_8 =\dfrac{1}{2} \times 8 \times (6 + 7d)$

$S_5 =\dfrac{1}{2} \times 5 \times (6 + 4d)$

So, using the given fact that $S_8 = 2S_5$, we see that

$\dfrac{1}{2} \times 8 \times (6 + 7d)=2 \times \dfrac{1}{2} \times 5 \times (6 + 4d)$

$24 + 28d = 30 + 20d$

$8d = 6$

$4d = 3$

$d = \dfrac{3}{4}$

### 4.4 Geometric Progressions

We shall now move on to the other type of sequence we want to explore. Consider the sequence

$2, 6, 18, 54, ... .$

Here, each term in the sequence is 3 times the previous term. And in the sequence

$1, -2, 4, -8, ... ,$

each term is -2 times the previous term. Sequences such as these are called geometric progressions, or GPs for short.Let us write down a general geometric progression, using algebra. We shall take a to be the first term, as we did with arithmetic progressions. But here, there is no common difference.

Instead there is a common ratio, as the ratio of successive terms is always constant. So we shall let $r$ be this common ratio. With this notation, the general geometric progression can be expressed as

$a, ar^2, ar^3....$

So the n-th can be calculated quite easily. It is $ar ^n-1$, where the power $(n - 1)$ is always one less than the position $n$ of the term in the sequence. In our first example, we had$a = 2$ and $r = 3$, so we could write the first sequence as

$2, 2 \times 3, 2 \times 3^2, 2 \times 3^3... .$

In our second example, $a = 1$ and$r = -2$, so that we could write it as

$1, 1 \times (-2), 1 \times (-2)^2, 1 \times (-2)^3 ......$

### 4.5 The Sum of a Geometric Series

Suppose that we want to find the sum of the first n terms of a geometric progression. What we get is:

$S_n = a + ar + ar^2+ar^3...ar^{n-1}$

and this is called a geometric series. Now the trick here to find the sum is to multiply by $r$ and then subtract:

$S_n = a + ar + ar^2+ar^3+ar^{n-1}$

$rS_n = ar + ar^2 + ar^3+ar^4+ar^n$

$S_n-rS_n = a -ar^n$

$S_n(1 - r) = a(1 -r^n)$

$S_n = \dfrac{a(1 -r^n)}{(1 - r)}$

#### Example:

Find the sum of the geometric series

$2 + 6 + 18 + 54 + ...$

where there are 6 terms in the series.

#### Solution:

For this series, we have:

$a = 2$,$r = 3$ and $n = 6$. So

$S_n(1 - r) = a(1 - r^n)$

$S_6(1-3)=2(1-3^6)$

$S_6=728$

#### Example:

Find the sum of the geometric series

$8 - 4 + 2 - 1 + ...$

where there are 5 terms in the series

#### Solution:

For this series, we have $a = 8$,$r = -1/2$ ,and $n = 5$. So

$S_n(1 - r) = a(1 - r^n)$

$S_5 =\dfrac{8(1-(-1/2)^5)}{(1-(-1/2))}$

$S_5 =\dfrac{8(1-(-1/32))}{(3/2)}$

$S_5=5\dfrac{1}{2}$

#### Example:

How many terms are there in the geometric progression

$2, 4, 8, ... , 128?$

#### Solution:

In this sequence $a = 2$ and $r = 2$. We also know that the n-th term is 128. But the formula for

the n-th term is $ar^{n-1}$

so $128 = 2 \times 2^{n-1}$

$64=2 \times 2^{n-1}$

$6 = n - 1$

$n = 7$ .

So there are 7 terms in this geometric progression.

#### Example:

How many terms in the geometric progression

$1, 1.1, 1.21, 1.331, ...$

will be needed so that the sum of the first $n$ terms is greater than 20?

#### Solution:

The sequence is a geometric progression with $a = 1$ and $r = 1.1$. We want to find the smallest value of n such that $S_n> 20$. Now

$S_n = \dfrac{a(1 - r^n)}{(1 - r)}$

$S_n = \dfrac{(1 - 1.1^n)}{(1 - 1.1)}$

$1.1^n-1>2$

$1.1^n>3$

If we now take logarithms of both sides, we get

$n \times \ln(1.1) > \ln(3)$

and as $\ln(1.1) > 0$ we obtain

$n >\dfrac{\ln(3)}{\ln(1.1)} = 11.5267$

and therefore the smallest whole number value of $n$ is 12

### 4.6 Convergence of Geometric Series

Consider the geometric progression

$1,\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16}, ...$

We have $a = 1$ and $r =1/2$ , and so we can calculate some sums. We get:

Table below can be scrolled horizontally

 $S_1$ $1$ $1$ $S_2$ $1+\dfrac{1}{2}$ $\dfrac{3}{2}$ $S_3$ $1+\dfrac{1}{2}+\dfrac{1}{4}$ $\dfrac{7}{4}$ $S_4$ $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}$ $\dfrac{15}{8}$

And there seems to be a pattern because

$1=2-1$

$\dfrac{3}{2}=2-\dfrac{1}{2}$

$\dfrac{7}{4}=2-\dfrac{1}{4}$

$\dfrac{15}{8}=2-\dfrac{1}{8}$

In each case, we subtract a small quantity from 2, and as we take successive sums. The quantity gets smaller and smaller. If we were able to add â€˜infinitely manyâ€™ terms, then the answer â€˜ought to beâ€™ 2 - or as near as we want to get to 2.

Let us see if we can explain this by using some algebra. We know that

$$S_n =\dfrac{a(1 -r^n)}{1 - r}$$

and we want to examine this formula in the case of our particular example where $r =\dfrac{1}{2}$

Now the formula contains the term $r^n$ and, as $-1 $$S_{\infty} =\dfrac{a}{(1 - r)}$$ where we have omitted the term$r^n$. We say that this is the limit of the sums$S_n$as$n\$ 'tends to infinity'.

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