smart mathematicians.... i need help?

Question

can u please write the equation that describes the sequence and pls. explain how did you get it..

1. 1,3,5,7,...

2. -5,-2,1,4,...

3. 3,7,11,15,...

4. 1,2,4,8,...

5. x+2,2x+1,3x,...

6. -5,-9,-13,-17,...

7. 12,7,2,-3,-8,...

8. 1,1.2,1.4, 1.6,...

9. 1.6, 1.9, 2.2, 2.5,...

tnx in advance...

Answer 1

We are asked to write an equation, and no answers so far have any equation.

In each of the equations, we will write a function that will give us the nth term in the sequence. The general procedure is to note the relationship between the successive terms in the series. In the first one, the series goes up by 2 when n goes up by 1. So the formula will need a term 2n in it somehwere. Then we have to consider the initial value of the series, so that for n = 1 we get the correct result.

1. T = 2n - 1

The second series is going to need a 3n term somewhere, but the series is -5 for n = 1.

2. T = 3n - 8

The fifth series is a little interesting, in that we add x and subtract 1 for each successive term. That means we will need n(x - 1) in the equation, but we have to make the initial value work.

3. n(x - 1) + 3

Matt Z, are you THE Matt Z?

Answer 2

First see if the same number is added each time to get the next number. Like in #3, you keep adding 4 to a number to get the next number. That means the expression 4n (where n = the position of the term in the sequence) has to be in the equation somewhere. If a sub n = the nth term in the sequence, you know its equation is a sub n = 4n + something. To get something, put in 1 for n and the answer should be 3. But 4n = 4. So you have to add -1, or subtract 1.

Answer: a sub n = 4n - 1


You can use this method on most of the other ones.
The only one here that's different is #4 because these numbers are MULTIPLIED by the same thing, 2. That means you will have 2 to a power of (n + something). Since 2^0 = 1, you need to add -1 or subtract 1 from n.

Answer: a sub n = 2^(n-1)

Answer 3

1. 1,3,5,7,... = 9 (add 2 to each #)

2. -5,-2,1,4,... = 7 (add 3 to each #)

3. 3,7,11,15,... = 19 (add 4 to each #)

4. 1,2,4,8,... = 16 (added 1, then 2, then 4, then 8...doubling #)

5. x+2,2x+1,3x,... = 3 (not sure, doesn't appear to be sequential due to the x variable, unless x=1 and every answer = 3, so therefor -- 1+2 = 3, 2(1) + 1 = 3, and 3(1) = 3).

6. -5,-9,-13,-17,... = -21 (subtracting by 4 for each #)

7. 12,7,2,-3,-8,... = -13 (subtracting by 5)

8. 1,1.2,1.4, 1.6,... = 1.8 (adding .2 to each #)

9. 1.6, 1.9, 2.2, 2.5,... = (2.8 adding .3 to each #)

Answer 4

Assuming (n) to be any integer (like a whole number, but for both positive and negative numbers), these are the mathematical expressions for the related sequences.

(1) 2n - 1 (odd number sequence)

(2) 3n + 1 (starting with an n = - 2) or 3n - 5 (starting with n at 0)

(3) 4n + 3

(4) 2^n (2 to the power of n, n of course starting out as 0 since

2n^0 equals 1)

(5) n(x - 1) + 3 (starting with 0)

(6) -4n - 5 (starting with 0)

(7) -5n + 12 (starting with 0)

(8) 0.2n + 1 (starting with 0)

(9) 0.3n + 1.6 (starting with 0)

That should hopefully answer your question. If you don't understand how I received these answers, then ask a teacher for help. Otherwise, check back or e-mail me.

Answer 5

1. 1 3 5 7 all you do here is add 2
2. -5 -2 1 4 all you do here is add 3
3. 3 7 11 15 all you do here is add 4
4. 1 2 4 8 all you do here is multiply by 2
I'm not sure for 5
6. -5 -9 -13 -17 all you do here is -4
7. 12 7 2 -3 -8 all u do here is - 5
8. 1 1.2 1.4 1.6 all you do here is plus .2
91.6 1.9 2.2 2.5 all you do here is add .3

Answer 6

1. 1,3,5,7,... - > nth term is 1 + (n-1)(2)
2. -5,-2,1,4,... - > nth term is -5 + (n-1)(3)
...
...
basically, for any arithmetic progression with first term "a" and common difference "d", the nth term is [ a + (n-1)d ]
just remember this standard result, it's useful.

err maybe i'll do number 5 too, to illustrate to you how to really use the formula.
5. first term a = x + 2
common difference = 2x + 1 - (x + 2) = x - 1
check, 3x - (2x + 1) = x - 1
therefore the nth term is : (x + 2) + (n-1)(x-1)

Answer 7

These aren't hard. Look at the differences between the numbers, and you'll get most of them. For example, on the first one, the difference between any two numbers is 2, so the sequence would continue 9, 11, 13, etc.

Try them on your own. :)

Answer 8

plus 2s, plus 3s,plus 4s, double the numbers, wait shouldn't you do your own homework?