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Sequences and Series: Basic Examples

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Your first homework on sequences and series will likely be a hodge-podge of generic exercises, intended to help you become familiar, and comfortable, with the basic terminology and notation. The exercises usually look scarier than they actually are. Give yourself some time, and work slowly through the problem set, so you can absorb the information you'll need later on.

Given the sequence An = {1, 3, 5, 7, 9},
(a) what is the value of a3?

(b) Find the value of \mathbf{\color{green}{\displaystyle{\sum_{\mathit{n}=1}^5\,\mathit{a}_{\mathit{n}}}}}
n=1
∑
5
​
a
n
​

Content Continues Below

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(a) The index of a3 is n = 3 so they're asking me for the third term, which is "5".

(b) The funky symbol is the Greek capital letter "sigma", indicating a series. That means that they're asking me here to do the addition of the terms of the sequence. The "value" they're asking me to find is the total, the sum, of all the terms an from a1 to a5; in other words:

a1 + a2 + a3 + a4 + a5

= 1 + 3 + 5 + 7 + 9 = 25

Then my answers are:

value of a3: 5

value of sum: 25

Expand the following series and find the sum: \mathbf{\color{green}{\displaystyle{ \sum_{\mathit{n}=0}^4\, 2\mathit{n} }}}
n=0
∑
4
​
2n
They've given me a rule for each term of this series; the rule is to multiply the index by two. So, to find each term, I'll plug the value of n into the formula; namely, I'll take the index and multiply by two. I'll be starting with n = 0 and ending with n = 4. To find the series sum, I'll be adding all the terms, like this:

2(0) + 2(1) + 2(2) + 2(3) + 2(4)

= 0 + 2 + 4 + 6 + 8 = 20

List the first four terms of the sequence {an} = {n2}, starting with n = 1.
I'll just plug n into the formula, and simplify:

{a1, a2, a3, a4}

= {12, 22, 32, 42}

= {1, 4, 9, 16}

My answer is the simplified form of the sequence:

{1, 4, 9, 16}

List the first four terms of the following sequence, beginning with n = 0.
\mathbf{\color{green}{ A_{\mathit{n}} = \dfrac{(-1)^{\mathit{n}}}{(\mathit{n} + 1)!} }}A
n
​
=
(n+1)!
(−1)
n

​

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Sequences and series are often the first place students encounter this exclamation-mark notation. The notation doesn't indicate that the series is "emphatic" in some manner; instead, this is technical mathematical notation. It indicates that the terms of this summation involve factorials. (If you're not familiar with factorials, brush up now.)

A factorial symbol, k!, indicates that I need to find the product of all the whole numbers from 1 through k. The first few factorial values are:

1! = 1

2! = 1 × 2 = 2

3! = 1 × 2 × 3 = 6

4! = 1 × 2 × 3 × 4 = 24

(Your graphing calculator can probably find factorials for you. Look for an appropriate command, probably somewhere in a "Prob" or "Probability" submenu.)

I'll use these factorial values in my computations:

a_0 = \dfrac{(-1)^0}{(0 + 1)!} = \dfrac{1}{1!} = \dfrac{1}{1} = 1a
0
​
=
(0+1)!
(−1)
0

​
=
1!
1
​
=
1
1
​
=1
a_1 = \dfrac{(-1)^1}{(1 + 1)!} = \dfrac{-1}{2!} = -\dfrac{1}{2}a
1
​
=
(1+1)!
(−1)
1

​
=
2!
−1
​
=−
2
1
​

a_2 = \dfrac{(-1)^2}{(2 + 1)!} = \dfrac{1}{3!} = \dfrac{1}{6}a
2
​
=
(2+1)!
(−1)
2

​
=
3!
1
​
=
6
1
​

a_3 = \dfrac{(-1)^3}{(3 + 1)!} = \dfrac{-1}{4!} = -\dfrac{1}{24}a
3
​
=
(3+1)!
(−1)
3

​
=
4!
−1
​
=−
24
1
​

So the first four terms are:

\mathbf{\color{purple}{1,\, -\dfrac{1}{2},\, \dfrac{1}{6},\, -\dfrac{1}{24}}}1,−
2
1
​
,
6
1
​
,−
24
1
​

Notice how, in that last example above, raising the –1 to the power n made the signs alternate. This alternating pattern of signs crops up a lot, especially in calculus, so try to keep this "raising –1 to the power n" trick in mind.

Content Continues Below

Find the sum of the first six terms of An, where an = 2an–1 + an–2, a1 = 1, and a2 = 1.
This formula looks much worse than it really is; I just have to give myself some time, and dissect the formula carefully.

They gave me the values of the first two terms, and then they gave me a formula that says that each term (after the first two terms) is a sum formed from the previous two terms. At each stage, I'll be taking the previous term and multiplying it by two; to this, I'll be adding the term before that one. For instance, the third term will be twice the second term, plus the first term.

Plugging into this formula, I get:

a3 = 2a3–1 + a3–2

= 2a2 + a1

= 2(1) + (1)

= 2 + 1 = 3

a4 = 2a4–1 + a4–2

= 2a3 + a2

= 2(3) + (1)

= 6 + 1 = 7

a5 = 2a5–1 + a5–2

= 2a4 + a3

= 2(7) + (3)

= 14 + 3 = 17

a6 = 2a6–1 + a6–2

= 2a5 + a4

= 2(17) + (7)

= 34 + 7 = 41

Now that I've found the values of the third through the sixth terms, I can find the value of the series; the sum is:

1 + 1 + 3 + 7 + 17 + 41 = 70

Write the following series using summation notation, beginning with n = 1:
2 – 4 + 6 – 8 + 10

The first thing I have to do is figure out a relationship between n and the terms in the summation. This series is pretty easy, though: each term an is twice n, so there is clearly a "2n" in the formula. I also have the alternating sign.

If I multiply 2n by (–1)n, then I'll get –2, 4, –6, 8, –10, which is backwards (on the signs) from what I want. But I can switch the signs by throwing in one more factor of –1:

(–1)(–1)n = (–1)1(–1)n = (–1)n+1

So the formula for the n-th term is an = (–1)n+1 (2n). Since n starts at 1 and there are five terms, then the summation is:

\mathbf{\color{purple}{\displaystyle{ \sum_{\mathit{n}=1}^5\,(-1)^{\mathit{n}+1}\,(2\mathit{n}) }}}
n=1
∑
5
​
(−1)
n+1
(2n)
Write the following using summation notation:
\mathbf{\color{green}{ \dfrac{5}{6+3} + \dfrac{5}{7+3} + \dfrac{5}{8+3} + ... + \dfrac{5}{31+3} }}
6+3
5
​
+
7+3
5
​
+
8+3
5
​
+...+
31+3
5
​

The only thing that changes from one term to the next is one of the numbers in the denominator.

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(Note: If I "simplify" these fractions, I'll lose this information. Any time the terms of my sequence or series look oddly lumpy, I tend not to simplify those terms: that odd lumpiness almost certainly contains a hint of the pattern I need to find.)

The changing numbers, as a list, start off with 6, 7, and 8. This looks like counting, but starting with 6 instead of 1. Without any information to the contrary, I'll assume that this is the pattern.

But I need to relate these "counting" values to the counter, the index, n. For n = 1, the number is 6, or n + 5. For n = 2, the number is 7, which is also n + 5. Checking the pattern for n = 3, 3 + 5 = 8, which is the third number. Then the terms seems to be in the following pattern:

a_n = \dfrac{5}{(n+5) + 3}a
n
​
=
(n+5)+3
5
​

But how many terms are in the summation? The ellipsis (the "..." or "dot, dot, dot" in the middle) means that terms were omitted. How many terms? Now that I have the general pattern for the series terms, I can solve for the counter (that is, for the value of n) for the last term:

31 = n + 5

31 – 5 = n + 5 – 5

26 = n

This tells me that there are 26 terms in this summation, so the series, in summation notation, is:

\mathbf{\color{purple}{\displaystyle{ \sum_{\mathit{n}=1}^{26}\, \dfrac{5}{(\mathit{n} + 5) + 3} }}}
n=1
∑
26
​

(n+5)+3
5
​

If the fractional forms of the terms in the series above had been simplified, it would have been a lot harder to figure out a pattern. So it's usually best to leave the terms in the form provided, rather than reducing them, because reducing would remove the pattern that they're wanting you to see.

To be fair, though, unless the sequence is very simple or is presented in a very straightforward manner, it is entirely possible that you might find a "wrong" pattern. Don't let this bother you terribly much. The "right" pattern is just the one that the author had in mind when he wrote the exercise. Your pattern would be "wrong" only in that it is unexpected. But if you can present your work clearly and logically, you should be able to talk your way into getting at least partial credit for your answer.

Once you've learned the basic notation and terminology, you will likely quickly move on to the two common and straightforward sequence types, being arithmetic and geometric sequences.

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