--%>
Assignment Help - homework help

  Theory of Geometric Sequences

Geometric Sequences:

The geometric sequence is the sequence in which the ratio of consecutive terms is constant.

Common Ratio:

As this ratio is common to all the consecutive pairs of terms, it is termed as the common ratio. It is represented by r. When the ratio among consecutive terms is not constant, then sequence is not geometric.

The formula for common ratio of a geometric sequence is r = an+1/an

General Term:

The geometric sequence is the exponential function. Rather than y = ax, we write an = crn here r is the common ratio and c is the constant (that is, not the first term of sequence, though).

The recursive definition, as each term is determined by multiplying the prior term by the common ratio, ak+1=ak * r.

The basic idea here is similar to that of arithmetic sequence, apart from each term is multiplied by the additional factor of r. The exponent on r will be one less than the term number. The first term has not been multiplied by r at all (that is, exponent on r is 0). The second term has been multiplied by r one time. The third term has been multiplied by r twice, and so forth.

The formula for general term of a geometric sequence is an = a1 rn-1

Partial Sum:

The series is a sum of a sequence. We wish to determine the nth partial sum or the sum of first n terms of the sequence. We will symbolize the nth partial sum as Sn.

Let consider the geometric series S5 = 2 + 6 + 18 + 54 + 162. There is a trick which can be utilized to determine the sum of series. In this series, r = 3.

S5 = 2 + 6 + 18 + 54 + 162

Multiply both sides of equation by -r = -3

-3 S5 = - 6 - 18 - 54 - 162 - 486

Now add up the two equations altogether. Note that whenever you do that, all however the first and last terms cancel out.

S5 - 3S5 = 2 - 486

We can factor a Sn out of the left hand side and factor a 2 out of right hand side.

(1-3) S5 = 2 (1 - 243)
Now divide both the sides by 1-3

S5 = 2 (1 - 243)/(1 - 3)

On simplifying it we get 2 (-242)/(-2) = 242. Well, whenever you are trying to make a conjecture regarding the general case, it does not generally help to simplify and drop the original values.

Now, when we try to figure out where the various parts of that formula come from, we can conjecture regarding a formula for the nth partial sum. The 2 in numerator is the first term, a1. The 243 in numerator is the ratio times the nth term - which makes it n+1 term, a1*rn. As both terms in numerator contain an a1 in them, which can be factored out. The 1 in denominator is always 1 and 3 in denominator was the ratio, r.

That forms the sum of first n terms Sn = a1 (1-rn)/(1-r).

There is implied domain which r can’t equivalent to the 1, however as it is implied; it doesn’t require to be stated.
The formula for nth partial sum of a geometric series is Sn = a1 (1-rn)/(1-r)

Infinite Sum:

There is the other kind of geometric series and infinite geometric series. The infinite geometric series is the sum of infinite geometric sequence.

Whenever the ratio consists of a magnitude more than 1, then the terms in the sequence will get bigger and bigger and if you add up bigger and bigger numbers forever, we will get infinity for an outcome. Therefore, we do not deal with infinite geometric series whenever the magnitude of the ratio is more than one.

Magnitude of the ratio cannot equivalent to one as that the series would not be geometric and the sum formula would encompass division by zero.

The only situation left, then, is whenever the magnitude of the ratio is less than one. Let consider r = 1/2. The sequence might be 1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128, 1/256, 1/512, 1/1024, 1/2048, 1/4096, 1/8192, 1/16384, 1/32768, 1/65536.... Since the sequence goes on, the terms are getting smaller and smaller, approaching to zero.

We know that a geometric sequence was an exponential function. What occurs to an exponential function if the base is between 0 and 1? This was asymptotic to the x-axis (y = 0) to the right. That is, since x approached infinity, y approached to 0. Well, the similar thing occurs here, as n approaches infinity, rn approaches to 0. Therefore, if you substitute rn with 0 in the summation formula, then the 1-rn part just becomes 1, and the numerator becomes a1.

The formula for the sum of infinite geometric series is S∞ = a1/(1-r).

Latest technology based Algebra Online Tutoring Assistance

Tutors, at the www.tutorsglobe.com, take pledge to provide full satisfaction and assurance in Algebra help via online tutoring. Students are getting 100% satisfaction by online tutors across the globe. Here you can get homework help for Algebra, project ideas and tutorials. We provide email based Algebra help. You can join us to ask queries 24x7 with live, experienced and qualified online tutors specialized in Algebra. Through Online Tutoring, you would be able to complete your homework or assignments at your home. Tutors at the TutorsGlobe are committed to provide the best quality online tutoring assistance for Algebra Homework help and assignment help services. They use their experience, as they have solved thousands of the Algebra assignments, which may help you to solve your complex issues of Algebra. TutorsGlobe assure for the best quality compliance to your homework. Compromise with quality is not in our dictionary. If we feel that we are not able to provide the homework help as per the deadline or given instruction by the student, we refund the money of the student without any delay.