Sequences and Summation Notation:
The sequence is a function whose domain is natural numbers. Rather than using the f(x) notation, though, a sequence is listed by using a notation. There are infinite sequences whose domain is the set of all the positive integers, and there are limited sequences whose domain is the set of first n positive integers.
Whenever you define a sequence, you should write the general term (nth term or an). There is sometimes more than one sequence which is possible when merely the first few terms are given.
Defining a Sequence:
There are two general ways to define a sequence by specifying the common term.
General Term, an:The first is to utilize a form which only depends on the number of term, n. To determine the first five terms if you know the general term, simply replace the values 1, 2, 3, 4 and 5 to the general form for n and simplify it.
Let consider the sequence defined by general term an = 3n-2
The first five terms are determined by plugging in 1, 2, 3, 4 and 5 for n.
3(1) - 2 = 13(2) - 2 = 43(3) - 2 = 73(4) - 2 = 103(5) - 2 = 13
Thus, the first five terms of sequence are 1, 4, 7, 10 and 13
Now consider the sequence stated by the general term an = 1/n.
The first five terms are 1/1, 1/2, 1/3, 1/4 and 1/5.
The second method is to recursively define a sequence. The recursive definition utilizes the current and/or preceding terms to define the next term. We can think of ak+1 being the next term, ak being the present term, and ak-1 being the prior term.
Let consider the sequence where a1 = 5 and ak+1 = 2 ak - 1.
We can read that last portion as ‘the next term is one less than twice the present term’.
The first five terms are as:
a) 5 (by definition),b) 2(5) - 1 = 9 (that is, twice the first term of 5 minus 1),c) 2(9) - 1 = 17 (that is, twice the second term of 9 minus 1),d) 2(17) - 1 = 33 (that is, twice the third term of 17 minus 1),e) 2(33) - 1 = 62 (that is, twice the fourth term of 33 minus 1)
Now consider the sequence stated by a1 = 2, a2 = 1, and ak+2 = 3ak - ak+1
We can read that last portion as ‘the next term is 3 times the last term minus the present term’.
First five terms are as:
a) 2 (by definition),
b) 1 (by definition),
c) 3(2) - 1 = 5 (that is, 3 times first term minus second term), Note that if k = 1, then the sequence gets written as a1+2 = 3a1 - a1+1 that becomes a3 = 3a1 - a2. As a1 = 2 and a2 = 1, this is where 3(2) - 1 = 5 comes from.
d) 3(1) - 5 = -2 (that is, 3 times second term minus the third term),
e) 3(5) - (-2) = 17 (that is, 3 times third term minus the fourth term)Fibonacci Series:
One of famous illustration of a recursively defined sequence is the Fibonacci series. The first two terms of Fibonacci series are 1 by definition. Each and every term subsequent to that is the sum of two preceding terms.
The Fibonacci series is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.... an+1 = an + an-1.
Fibonacci series take place frequently in nature. For illustration, take a leaf on stem of many plants (such as cherry, elm or pear trees). Count the number of leaves till you reach one directly in line with one you chosen. The total number of leaves (not comprising the first one) is generally a Fibonacci number. When the left and right handed spirals on pine cone, sunflower seed heads or the pineapples are counted, then the two numbers are frequently consecutive Fibonacci numbers.Factorials - !
The factorial is denoted by! (an explanation point). Factorial of a positive integer is the product of all positive integers less than or equivalent to that number. Zero factorial is a special case and 0! = 1 by definition.
1! = 12! = 2*1! = 2*1 = 23! = 3*2! = 3*2*1 = 64! = 4*3! = 4*3*2*1 = 245! = 5*4! = 5*4*3*2*1 = 120n! = n(n-1)! = n*(n-1)*(n-2)*...3*2*1
Note that there is also a recursive definition there. Any number factorial is that number times the factorial of one less than that number.
On TI-82 and TI-83 calculators, the factorial key can be found beneath Math, Probability, and menu choice 4.
Simplifying ratios of factorials:
Let consider 8!/5!. One manner to work the problem would be to completely expand the 8! and completely expand the 5!.
It is noted that there are common factors of 5, 4, 3, 2 and 1 in both numerator and denominator which divide out. This leaves you with 8*7*6 in the numerator for an outcome of 336.
The other way to work the problem; though is to use the recursive nature of factorials. As any number factorial is that number times the factorial of one less than number, 8! = 8 * 7!, however 7! = 7 * 6!, and 6! = 6 * 5!. This signifies that 8! = 8 * 7 * 6 * 5!. Therefore, 8!/5! Is 8*7*6*5!/5! = 8*7*6 = 336.
The main point is that there is no requirement to multiply the whole thing out whenever you are just going to divide a bunch of it out anyhow.
Here are a few steps to simplifying the ratio of the two factorials.
a) Find out the smaller factorial and write it down. It won't modify, and it will be the portion that is totally divided out.
b) Take the bigger factorial and begin expanding it by subtracting one till the smaller number (which you have already written down) is reached.
c) Whenever you reach the smaller number, write it as the factorial and divide out the two equivalent factorials.
Here is an illustration with a little bit more complex ratio. As 2n-1 is less than 2n+1, then copy the 2n-1 factorial down on bottom. Now, take 2n+1 times one less than 2n+1. One less then 2n+1 is 2n+1-1 = 2n. Take that times one less than 2n, that is 2n-1. As that 2n-1 is the similar amount on the bottom, call 2n-1 factorial and then divide it out, leaving the product of 2n and 2n+1 as an outcome.
Summation is something which is completed quite frequently in mathematics, and there is a symbol which signifies summation. That symbol is a capital Greek letter sigma, and therefore the notation is sometimes termed as Sigma Notation rather than Summation Notation.
The k is termed as the index of summation. k =1 is the lower limit of summation and k = n (though the k is only written once) is the upper limit of summation.
What the summation notation signifies is to compute the argument of the summation for each and every value of the index among the lower limit and upper limit (inclusively) and then add up the results altogether.
Replace each value of k between 1 and 5 to the expression 3k-2 and then add the results altogether.
[3(1)-2] + [3(2)-2] + [3(3)-2] + [3(4)-2] + [3(5)-2] = 1 + 4 + 7 + 10 + 13 = 35
[3(1) + 3(2) + 3(3) + 3(4) + 3(5)] - 2 = [3 + 6 + 9 + 12 + 15] - 2 = 45 - 2 = 43
Though such look much similar, the answers are distinct. The second illustration is thrown in there to warn you regarding notation. Multiplication consists of a higher order of operations than subtraction or addition, therefore no group symbols are required around 3k. However as subtraction has similar precedence as addition, the subtraction of 2 doesn’t go within the summation. In another word, be sure to comprise parentheses around a sum or difference when you wish for the summation to apply to more than just first term.
Properties of Summation:
The given below properties of summation apply no matter what the lower and upper limits are for index. For simplicity sake, we will not write k = 1 and n, however know that your index of summation is k in the given illustrations.
We can factor a constant out of a sum:
∑cak = c∑ak
Note ak consists of a subscript of k whereas the c does not. This signifies that c is a constant and a is function of k. The sum of constant times a function is the constant times the sum of function.
The sum of a sum is the sum of the sums:
It just sounds good. The summation symbol can be distributed over the addition.
∑(ak + bk) = ∑ak + ∑bk
The sum of a difference is the difference of the sums:
The summation symbol can be distributed over the subtraction.
∑(ak - bk) = ∑ak - ∑bk
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