Some couple of questions

EbenezerSon · 2015-12-27 22:33:59

1) Determine the next three terms of the sequence ,1,3,6,10,
And my attempts as follows;
common difference, 3 - 1 = 2
firt term = 1, second term 1 + 2 = 3, third term 1 +2(2) = 5, fourth term 1 + 3(2) = 7, fith term 1 + 4(2) = 9, sixth term 1 + 5(2) = 11 and so on. Is these correct?

Again,
(2)Find the coefficient of X^2 in the expression (1 + x )^4.
My solution is [1, 4, 6, 4, 1] is that correct?

(3)Find the sum of the sequence of the natural numbers 1, 2, 3, 4,........25.
How are going to do this?

4) Write down the next three terms of the sequence 12, 18, ,27 , ,

Thanks!

Relentless · 2015-12-27 22:51:56

The sequence in 1. is not arithmetic or geometric; there is no common difference or common ratio. The pattern for 1. seems to be 1 + 2 + 3 + 4... These numbers are called triangular numbers, and I think the Pythagoreans revered them over 2500 years ago. There is a polynomial rule for this, it is

You can also use the above rule for Q3

For 2., you are correct.

Last edited by Relentless (2015-12-28 02:38:33)

Bob · 2015-12-27 22:56:06

hi Relentless,

I was just about to post here when I saw you were already 'on the case'. Q4 looks odd. How are we to determine from so few terms. And is that double comma indicative of a missing term ?

Bob

Relentless · 2015-12-27 23:05:00

Hi Bob

I'm not sure whether there is supposed to be an additional term, but I don't think there is another term between 18 and 27 despite that double comma. There aren't many terms, and there might be several rules, but once again it looks like there is a fairly simple polynomial that will fit.

EDIT: No I think that is how it is supposed to be. The pattern is (9 + 3) + 6 + 9 + 12... The polynomial for that is

Last edited by Relentless (2015-12-27 23:14:39)

Relentless · 2015-12-27 23:28:10

A final note: A more intuitive way of solving 3. than just using the formula is to do what the "Prince of Mathematicians", Carl Gauss, apparently did as a schoolboy. He was told to write out the sum of natural numbers up to 100, and quickly solved it by recognising that most of the numbers could be matched up in pairs that add up to 100, for example 100+0=100, 99+1=100, 98+2=100... right down to 51+49 = 100, counting fifty pairs plus 50, giving a speedy answer of 5050.

The same can be done with sums up to 25, or any amount. With 25+0=25, 24+1=25, 23+2=25... you will promptly get to 13+12=25 and count thirteen pairs.

Last edited by Relentless (2015-12-27 23:29:26)

Bob · 2015-12-27 23:40:00

12 = 2 x 2 x 3
18 = 2 x 3 x 3
27 = 1 x 3 x 3 x 3

So the general rule could be

This would make the next three terms:

0.5 x 81
0.25 x 243
0.125 x 729

That's the trouble when there are few terms given and no other clues.

Bob

Relentless · 2015-12-27 23:54:13

That is trouble. A graphing calculator gave me an exponential that fits precisely as well. I suppose one rule will seem more implicit than another, but an open question shouldn't require a closed answer.

Wouldn't it be theoretically possible to construct several different (perhaps monstrously complex) formulas to fit the same sequence in many, if not most or all, cases?

Last edited by Relentless (2015-12-28 02:57:33)

phanthanhtom · 2015-12-28 01:51:34

That's right.

A set of n values always defines exactly one polynomial of degree n-1 (or less), and infinite polynomials of degrees n, or n+1, or n+2 etc.

So in general, if you are given something like

1, 3, 5, 7...

you'll find one matching <4 polynomial, and infinite matching degree n (n>3) polynomials for each n. To find that degree 3 polynomial, simply plug in the values x = 1, 2, 3, 4 into P(x) = ax^3 + bx^2 + cx + d (you'll get a system of 4 degree 1 equations with 4 unknowns, so it will have exactly one solution). Of course there will be special cases, but in general, if you have n given numbers in the sequence, there would be infinite formulas that fit unless you limit the function to be a polynomial function of degree not exceeding n-1.

Something that goes with my previous post: http://mth.bz/fit/polyfit/findcurv.htm

Last edited by phanthanhtom (2015-12-29 01:30:24)

Relentless · 2015-12-28 02:35:00

Thank you It's really not too difficult once it's explained, at least for polynomials.
I solved that system of equations and got a=b=0. I take that to mean that there is one solution of degree n-1 OR less?

EbenezerSon · 2015-12-28 07:56:17

Hi Relentless,

How did you use the rule 0.5n^2 + 0.5n to answer question one? Please show working

Last edited by EbenezerSon (2015-12-28 07:57:24)

Relentless · 2015-12-28 19:30:49

Oh! Sorry I didn't reply sooner, it has been a busy day.

As phanthanhtom mentioned, you find the rule for an n-1 degree polynomial (where n is the number of terms you have), and plug x = 1, 2, 3, ... n into the polynomial a + bx + cx^2 + ... zx^(n-1) and set it equal to term 1, 2, 3, ... n respectively to solve for n coefficients with n equations.
Honestly, I mostly just use a graphing calculator for this, but I will attempt to run you through it with question 1.

We have four terms, so our polynomial will be of degree 4-1 = 3. The general polynomial of degree 3 is ax^3 + bx^2 + cx + d. Our job is to find out a,b,c,d, and that is our rule. The four equations we get with our four terms are:
x = 1, so a + b + c + d = 1
x = 2, a*8 + b*4 + c*2 + d = 3
x = 3, a*27 + b*9 + c*3 + d = 6
x = 4, a*64 + b*16 + c*4 + d = 10

There are lots of ways to proceed from here, and somebody else probably has a much faster way, but if we substitute 1-b-c-d for a, we get:
4b + 6c + 7d = 5
18b + 24c + 26d = 21
48b + 60c + 63d = 54

Substituting -1.5c - 1.75d + 1.25 for b, we get:
3c + 5.5d = 1.5
12c + 21d = 6

Substituting 0.5 - 11d/6 for c, we finally get: d = 0
Substituting d = 0, we get c = 0.5
Substituting both of those, we get b = 0.5
Substituting all of those, we get a = 0

The final rule is y = 0.5x^2 + 0.5x

Last edited by Relentless (2015-12-28 20:59:53)

Relentless · 2015-12-29 04:04:57

As for how it was used, once you have a rule you just make n equal the term number you want. So, 0.5(4)^2 + 0.5(4) will give you 10, the fourth term. The fifth is 0.5(5)^2 + 0.5(5) = 15, sixth 0.5(6)^2 + 0.5(6) = 21, seventh 0.5(7)^2 + 0.5(7) = 28, and the 342nd is 0.5(342)^2 + 0.5(342) = 58,653

EbenezerSon · 2015-12-29 08:24:50

So, one has to generate his own rule by using the numbers given? Or that rule you have given should always be used for such questions?

Again,
So, if the numbers have no common difference or ratio then I should resort to using the polynomials like the rule you have generated for the problem?

Last edited by EbenezerSon (2015-12-29 08:26:32)

Relentless · 2015-12-29 18:33:51

Hi EbenezerSon,

That is a systematic way of doing it if you want to be able to give an answer for any number in the sequence, even the, say 5,000th. In this case, you only want the next few terms, so if you can spot any way of connecting the numbers, that will do. For question 1, I realised that they were triangle numbers, i.e. first 1, then 1+2, then 1+2+3, then 1+2+3+4, and so on. It turns out that this pattern is exactly the same as the polynomial, just more simple to explain.

There also may be several ways of connecting the numbers, as bob pointed out with question four. Again, I spotted that the first is 9+3, second 9+3+6, third 9+3+6+9, and so on. Which, again, is the same rule as the polynomial. But bob gave 2^(3-n) * 3^n, which also works, but defines a totally different sequence.

There is often more than one way to go about it, and pattern recognition is very useful. But being the systematic type, yes, I tend to just make a polynomial. (:

phanthanhtom · 2016-01-04 21:23:05

Oh, and I have this to add:

The reason why there is only one polynomial solution (i.e. the polynomial associated with the sequence) for degrees <= n-1 if you have n numbers given is because it would then be reduced to a system of n linear equations with n unknowns.

However, in theory, there are other ways to reduce the question to such a system.

Take your sequence 1, 3, 6, 10. You can (at least in theory) interpret the sequence as 2^(f(x)), 3^(f(x)) etc., where f(x) would be a polynomial with degree <= 3 to be determined.

You would then have f(1) = log(3)1 = 0, f(2) = log(3)3 = 1, f(3) = log(3)6 = 1 + log(3)2, f(4) = log(3)10. Then remember that f(x) = ax^3 + bx^2 + cx + d and you would, in theory, be able to find the coefficients a, b, c and d.

This would be easier to show for another sequence, such as, 3, 3, 3, 9... The correct degree 3 polynomial is x^3 - 6x^2 + 11x - 3. However, the sequence could also be interpreted as f(x) = 3^g(x), where g(x) = (1/6)(x-1)(x-2)(x-3) + 1. In the first case, the next number would be 27; in the second case, the next number would be 243.

Have fun with all these sequences! Try finding as many possible functions as possible.

Last edited by phanthanhtom (2016-01-04 21:23:22)

EbenezerSon · 2016-01-31 01:48:28

Hi, Please help with these;
It will help me understand any question regarding G.P and A.P

(1) The sum of the first three terms of a geometric sequence is 8 and the sum of the first six terms is 12. Find the the common ratio

(2) The sum of the first twelve of an A.P is 168. If the term is 7, find the values of
(i) the common difference
(ii) the first term

(3) the sum of the first 3 terms of a G.P is 14. If the term is 7, find the possible values of the sum of the first 5 terms

Please help me I got an exam by tomorrow many thanks.

Nehushtan · 2016-01-31 03:06:04

(1)

Substitute from [1] and solve.

EbenezerSon · 2016-01-31 07:53:35

Thank Nehu!

Please-- how about the others?
Please help.

Nehushtan · 2016-01-31 08:16:53

EbenezerSon wrote:

Please-- how about the others?

How about you putting in some of your own effort?

EbenezerSon · 2016-01-31 08:23:45

Just question three [3]
the answer is at the back of the book and is 62 or 122 but need the steps

bobbym · 2016-01-31 11:45:59

If the term is 7

What term?

Nehushtan · 2016-01-31 21:21:16

EbenezerSon wrote:

Just question three [3]
the answer is at the back of the book and is 62 or 122 but need the steps

I had to work backwards from the answers to figure out the mistake in your post.

EbenezerSon wrote:

(3) the sum of the first 3 terms of a G.P is 14. If the first term is 2, find the possible values of the sum of the first 5 terms

If you seriously want help, you have to make an effort yourself. How is anyone to help you if you don't bother to post your question correctly? I was able to work out your typo because I happened to have time to do that. You must bear in mind that most people are busy and have things to do in life, so they won't be as blessed with free time as I am!

Anyway.

Now you carry on from there.

Last edited by Nehushtan (2016-02-01 04:12:55)

Math Is Fun Forum

#1 2015-12-27 22:33:59

Some couple of questions

#2 2015-12-27 22:51:56

Re: Some couple of questions

#3 2015-12-27 22:56:06

Re: Some couple of questions

#4 2015-12-27 23:05:00

Re: Some couple of questions

#5 2015-12-27 23:28:10

Re: Some couple of questions

#6 2015-12-27 23:40:00

Re: Some couple of questions

#7 2015-12-27 23:54:13

Re: Some couple of questions

#8 2015-12-28 01:51:34

Re: Some couple of questions

#9 2015-12-28 02:35:00

Re: Some couple of questions

#10 2015-12-28 07:56:17

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#11 2015-12-28 19:30:49

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#12 2015-12-29 04:04:57

Re: Some couple of questions

#13 2015-12-29 08:24:50

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#14 2015-12-29 18:33:51

Re: Some couple of questions

#15 2016-01-04 21:23:05

Re: Some couple of questions

#16 2016-01-31 01:48:28

Re: Some couple of questions

#17 2016-01-31 03:06:04

Re: Some couple of questions

#18 2016-01-31 07:53:35

Re: Some couple of questions

#19 2016-01-31 08:16:53

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#20 2016-01-31 08:23:45

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