sequences and series

A sequence is a set of terms which are written in a definite order obeying certain rules e.g. 2,4,6,8,10,... is an infinite (since it goes on and on forever) sequence obeying a certain rule which is; for you to get the next term in the sequence you add 2 to the preceding one.In particular this sequence is made of multiples of two or even better the even numbers.

A sequence with definite terms is said to be a finite sequence e.g. 3,6,9,12. Note that they are no three dots after 12 i.e. the three dots normally denote that the sequence goes on and on forever. Hence this sequence has 4 terms. Hence  it is a finite sequence.

A series is the sum of the terms of a sequence e.g. 2+4+6+8+10+... is a series. In general a finite series is the sum of terms of a finite sequence i.e. a1+a2+a3+...+an. While an infinite series is the sum of terms of an infinite sequence i.e. a1+a2+a3+...+an+....

The most common sequences are the arithmetic sequences and the geometric sequences. An arithmetic sequence is a sequence that proceeds with a common difference which is normally denoted by d.In general an arithmetic sequence is a sequence of the form; a, [a+d], [a+2d],[a+3d],...,[a+(n-1)d] where a is the first term, n  is the number of terms in the sequence and [a+(n-1)d] is the last term which is either denoted as l or Tn (i.e. the nth term). For example; 2,4,6,8,10,... is an arithmetic sequence with first term a=2 and common difference d=2.

An Arithmetic series also referred to as an Arithmetic progression is the sum of terms of an arithmetic sequence i.e. a+[a+d]+[a+2d]+[a+3d]+...+[a+(n-1)d] e.g. 4+7+10+13+...is an infinite arithmetic series with first term a=4 and common difference d=3


On the other hand a geometric sequence is a sequence that proceeds with a common ratio normally denoted by r .In general a geometric sequence is a sequence of the form a,ar,ar^2,ar^3,...,ar^(n-1),where a is the first term, r is the common ratio,n is the number of terms in the sequence and ar^(n-1) is the last term or the nth term of the sequence. For example; 2,4,8,16,32,... is a geometric sequence with first term a=2, and common ratio r=2

A geometric series or the geometric progression is the sum of terms of a geometric sequence i.e. a+ar+ar^2+ar^3+...+ar^(n-1) e.g. 3+9+27+81+...is an infinite geometric series with first term a=3 and common ratio r=3.

Solving a quadratic equation by completing the square method

I will demonstrate by two examples. Example 1 has the coefficient of x^2=1 and Example 2 has the coefficient of x^2 greater than 1
Example 1
solve by completing the square method; x^2-7x+10=0 (note also the coefficient of x =-7)
solution
The first step is to take the constant 10 to the RHS to get;
x^2-7x=-10
Next we add a constant k to both sides to get;
x^2-7x+k=-10+k ....(*)
Next we find the value of k to 'complete the square' on the LHS by making it a perfect square i.e.;
k=(1/2*the coefficient of x)^2
k=(1/2*-7)^2=(-7/2)^2
Now we write equation (*) as follows;
x^2-7x+(-7/2)^2=-10+(-7/2)^2
The LHS can be easily factorized as below;
(x-7/2)^2=-10+49/4
(x-7/2)^2=9/4
We next find the square root of both sides to get;
x-7/2=±3/2
x=7/2±3/2
x=3.5±1.5
x=5 or 2
Example 2
Solve by completing the square method; 2x^2+x-3=0 (Note that the coefficient of x^2=2 unlike example 1 above)
Our first step is to make the coefficient of x^2=1 by dividing every term by 2 to get;
x^2+x/2-3/2=0
Next we take the constant to the RHS to get;
x^2+x/2=3/2
Next we add a k to both sides to get;
x^2+x/2+k=3/2+k.....(**)
Next we find the value of k to 'complete the square' on the LHS by making it a perfect square i.e.;
k=(1/2*coefficient of x)^2=(1/2*1/2)^2=(1/4)^2
Equation (**) can now be written as;
x^2+x/2+(1/4)^2=3/2+(1/4)^2
the LHS can be factorized to get;
(x+1/4)^2=3/2+1/16
(x+1/4)^2=9/16
Next find the square of both sides to get;
x+1/4=±3/4
x=/1/4±3/4= 4/4 or -2/4 
x=1 or -1/2

Solving by a quadratic equation using the Quadratic Formula

Given a quadratic equation ax^2+bx+c=0 then x=(-b±√(b^2-4ac))/2a. This is the quadratic formula. Example 1 Solve the following equation using the quadratic formula; 3x^2+7x+2=0 solution. In our case a=3, b=7, and c=2. We next need to replace a, b,and c with 3, 7, and 2 respectively in the formula to get; x=(-7±√(7^2-4*3*2))/2*3. x=(-7±√(49-24))/6 = (-7±5)/6 = -2/6 or -12/6 =-1/3 or -2.

Solving a quadratic equation by factorization

A quadratic equation is an equation of the form ax^2+bx+c=0, where a, b, and c are known constants and a  i not zero e.g. 2x^2+3x+7=0; 3x^2-4x=0; 7x^2-9=0 etc.

To solve by factorization, first factorize the LHS i.e. ax^2+bx+c, by first finding two factors which i will call T1 and T2 such that; (i) their sum is b i.e. T1+T2=b and (ii) their product is ac i.e. T1*T2=ac. Next replace bx with T1x and T2x to obtain ax^2+T1x+T2x+c=0. Then carry out group factorization of the LHS.

I will next demonstrate using two examples;
Example 1
Solve by factorization; x^2+7x+12=0
Solution
In this example our a=1, b=7 and c=12. Hence we need two factors T1 and T2 such that;
(i) T1+T2=7 and (ii) T1*T2=12. The two factors are T1=3 and T2= 4.
Next we replace 7x with 3x and 4x to get;
x^2+3x+4x+12=0 ......(***)
The next step is to factorize the LHS by group factorization i.e. we take the first two terms x^2 and 3x and factorize [i.e. x^2+3x=x(x+3)] and then factorize the next two terms i.e. [4x+12=4(x+3)]
Such that our equation (***) above becomes;
x^2+3x+4x+12=x(x+3)+4(x+3)=0
Note that (x+3) is common thus we have;
(x+3)(x+4)=0
(x+3) and (x+4) are the two factors of x^2+7x+12. Since their product is 0 it implies that either;
x+3=0 or x+4=0
Hence x=-3 or x=-4.
-3 and -4 are the solutions or the roots of the quadratic equation x^2+7x+12=0
Example 2.
solve 3x^2+10x+8=0
Solution
In this example our a=3, b=10 and c=8
Hence we need two factors T1 and T2 such that (i) T1+T2=10 and (ii) T1*T2=24
The two factors are 4 and 6.

Next we replace 10x with 6x and 4x to get;

3x^2+6x+4x+8=0 ......(******)

The next step is to factorize the LHS by group factorization i.e. we take the first two terms 3x^2 and 6x and factorize [i.e.3x^2+6x=3x(x+2)] and then factorize the next two terms i.e. [4x+8=4(x+2)] 

Such that our equation (******) above becomes;

3x^2+6x+4x+8=3x(x+2)+4(x+2)=0

Note that (x+2) is common thus we have;

(x+2)(3x+4)=0

(x+2) and (3x+4) are the two factors of 3x^2+8x+10. Since their product is 0 it implies that either;

x+2=0 or 3x+4=0

Hence x=-2 or x=-4/3. 

-2 and -4/3 are the solutions or the roots of the quadratic equation 3x^2+8x+10=0