Maths FormulasWatch

#1
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13 years ago
#2
(Original post by alexsmithson)
Hi, does anyone know what the formulas are to these sequences?

1] 14,25,37,50,64
2] 22,34,46,58,70
3] 7,17,28,40,53
4] 17,29,42,56
5] 2,7,14
6] 2,6,11,17,24
7] 3,8,15,24,35
8] 17,24,31,38,45

I will give positive rep to anyone that can help.

Look at the differences between consecutive terms.

These questions are always a little absurd as there are infinitely many formulae that would work. #5, with only three terms, is especially daft.
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13 years ago
#3
(Original post by alexsmithson)
Do you know the answers though?

Thanks, Alex.
Surely you can work them out once you note the pattern in the differences. When the differences are no worse than linear then the terms themselves will be no worse than quadratic.
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13 years ago
#4
1] 14,25,37,50,64
2] 22,34,46,58,70
3] 7,17,28,40,53
4] 17,29,42,56
5] 2,7,14
6] 2,6,11,17,24
7] 3,8,15,24,35
8] 17,24,31,38,45
2) 12n + 10
5) n² + 2n - 1
8) 7n + 10

Try searching 'sequences and series' or something like that on google, I'm sure you'll be able to find out something about the method of differences ...

Hope this helps,

~~Simba
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13 years ago
#5
(Original post by alexsmithson)
Hi, does anyone know what the formulas are to these sequences?

1] 14,25,37,50,64
2] 22,34,46,58,70
3] 7,17,28,40,53
4] 17,29,42,56
5] 2,7,14
6] 2,6,11,17,24
7] 3,8,15,24,35
8] 17,24,31,38,45

I will give positive rep to anyone that can help.

1] 14,25,37,50,64
Differences: 11, 12, 13, 14..

differences are linear, so our function will be quadratic.

let tn = an² + bn + c

n = 0: t0 = c = 14.
n = 1: t1 = a + b + c = 25
n = 2: t2 = 4a + 2b + c = 37

a+b = 11
4a+2b = 23
2a = 1 -> a = ½ -> b = 10½

tn = ½(n² + 21n + 28)

2] 22,34,46,58,70
differences: 12, 12, 12 ,12

tn = 22 + 12n.

3] 7,17,28,40,53
differences: 10, 11, 12, 13
note the differences increase at the same rate as in 1], so we can cheat.
let a = sequence from Q1, b = sequence from this question.
b0 = a0 - 7.
b1 = a1 - 8.
b2 = a2 - 9.
b3 = a3 - 10 ...

bn = an - 7 - n.
= ½(n² + 21n + 28) - 7 - n = ½(n² + 19n + 14)

4] 17,29,42,56
differences:
12,13,14.

same method as 3], answer is:
tn = ½(n² + 23n + 34)

5] 2,7,14
This one really is a bit short. One idea is 2x7 = 14, and tn+2 = tn * tn+1. This will give you tn = 2^(Fn-2) * 7^(Fn-1) for n >= 2 (Fn is the nth fibonacci number). I don't think that's what they are looking for , so just assume it's like 1]...

n = 0: t0 = c = 2.
n = 1: t1 = a + b + c = 7
n = 2: t2 = 4a + 2b + c = 14

a + b = 5.
4a + 2b = 12.
-> 2a = 2.
-> a = 1,b = 4,c=2.

tn = n² + 4n + 2

6] 2,6,11,17,24
dif: 4,5,6,7

tn = ½(n² + 7n + 4)

7] 3,8,15,24,35
dif: 5,7,9,11

tn = n² + 4n + 3

8] 17,24,31,38,45
dif: 7,7,7,7

tn = 17 + 7n
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13 years ago
#6
Meathead has the right idea, but as far as I'm aware at GCSE level at least, and up to where I am at A-level, it's normal to start with t1 than t0.

So, both mine and Meathead's solutions are correct ...
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13 years ago
#7
(Original post by alexsmithson)
So far, I have got these.

1] 14,25,37,50,64-----=-----½(n²+21n+28) ----->is there any way to make this smaller?
2] 22,34,46,58,70-----=-----12n+10
3] 7,17,28,40,53-------=-----½(n²+19n+14) ----->is there any way to make this smaller?
4] 17,29,42,56---------=-----½(n²+23n+34) ----->isthere any way to make this smaller?
5] 2,7,14---------------=-----n²+2n-1
6] 2,6,11,17,24--------=-----½(n²+7n+4) ----->is there any way to make this smaller?
7] 3,8,15,24,35--------=-----n²+2n
8] 17,24,31,38,45-----=-----7n+10

Just wanted to make them all in the same form. Can it be done?

Thanks again, Alex.
they are of the same form - quadratics. just some of the coefficients are fractions.
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13 years ago
#8
(Original post by alexsmithson)
So far, I have got these.

1] 14,25,37,50,64-----=-----½(n²+21n+28) ----->is there any way to make this formula smaller?
2] 22,34,46,58,70-----=-----12n+10
3] 7,17,28,40,53-------=-----½(n²+19n+14) ----->is there any way to make this formula smaller?
4] 17,29,42,56---------=-----½(n²+23n+34) ----->isthere any way to make this formula smaller?
5] 2,7,14---------------=-----n²+2n-1
6] 2,6,11,17,24--------=-----½(n²+7n+4) ----->is there any way to make this formula smaller?
7] 3,8,15,24,35--------=-----n²+2n
8] 17,24,31,38,45-----=-----7n+10

Just wanted to make them all in the same form. Can it be done?

Thanks again, Alex.
The first,third,fourth and sixth, only work if u call the first term n=0 so try replacing n with n+1 so you will get the first term by putting in n=1.
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13 years ago
#9
(Original post by alexsmithson)
So would they be:

.
1] . . Formula = ½((n-1)²+21(n-1)+28)

.
3] . . Formula = ½((n-1)²+19(n-1)+14)

.
4] . . Formula = ½((n-1)²+23(n-1)+34)

.
6] . . Formula = ½((n-1)²+7(n-1)+4)

?????
yeah now try to see if u can simplify that...
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13 years ago
#10
.
1] . . Formula = ½((n-1)²+21(n-1)+28)
So, ½(n²-2n+1+21n-21+28)
=½(n²+19n+8)

.
3] . . Formula = ½((n-1)²+19(n-1)+14)
So, ½(n²-2n+1+19n-19+14)
=½(n²+17n-4)

Can't be bothered to do the rest, it's all simple stuff.
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