Cant get this math. - Printable Version
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Cant get this math. by Baggins on 04-27-2008 at 11:07 AM
I have to write expressions for patterns.
example:
Pattern: 8, 13, 18, 23, 18, 33
Description: Start at 8 then add 5.
Expression: n5+3
There are two patterns that I couldn't figure out, and nither could my teacher when I asked her.
1.
Pattern: 3, 6, 10, 15, 21
Description: Start at 3, then add 3, next add 4 and so on and so forth.
Expression: ???
2.
Pattern: 8, 16, 32, 64, 128
Description: Start at 8, then multiply by 2.
Expression: ???
Please help. 
RE: Cant get this math. by John Anderton on 04-27-2008 at 11:26 AM
1) Start at 3; Add 3, then 4 then 5...
q=p+(3+n)
Where q = current answer (ie nth value)
and p = previous answer (ie (n-1)th value)
Initialise n=0 and p=3
4) Indices of 2 starting from 8
q=p*2
Initialise p = 8
Or, q=2^(n+3) 
quote:More like 128
Originally posted by Bilbo
8, 16, 32, 64, 182
RE: Cant get this math. by Phillip on 04-27-2008 at 11:39 AM
quote:Wow your math teacher must be kinda stupid =\
Originally posted by Bilbo
There are two patterns that I couldn't figure out, and nither could my teacher when I asked her.
RE: Cant get this math. by Th3rmal on 04-27-2008 at 11:41 AM
quote:incredibly, if the teacher doesnt understand it, then why is she giving them these questions without knowing how to do it first?
Originally posted by Phillip
quote:Wow your math teacher must be
Originally posted by Bilbo
There are two patterns that I couldn't figure out, and nither could my teacher when I asked her.
kindastupid =\
RE: Cant get this math. by Baggins on 04-27-2008 at 11:43 AM
I'm still in grade 8, it's not like all the teacher does is math. In fact, she does everything but french.
EDIT:
JA: You're supposed to be able to do it with only n, without knowing the previous number.
RE: Cant get this math. by Th3rmal on 04-27-2008 at 11:47 AM
quote:you rock, but the teacher should know how to do the equations...
Originally posted by Bilbo
I'm still in grade 8, it's not like all the teacher does is math. In fact, she does everything but french.
RE: Cant get this math. by Baggins on 04-27-2008 at 11:59 AM
also, JA.
quote:it's q=2^(n+2)
Originally posted by JA
q=2^(n+3)
RE: Cant get this math. by markee on 04-27-2008 at 12:11 PM
1) p = (n+2)(n/2+1.5)
I realise it isn't the easiest to find, but it shouldn't take a maths teacher more than 5min... And where was the class nerd to anser this one?
RE: Cant get this math. by Baggins on 04-27-2008 at 12:22 PM
quote:Thanks, and just ftw I am the class nerd.
Originally posted by markee
1) p = (n+2)(n/2+1.5)
I realise it isn't the easiest to find, but it shouldn't take a maths teacher more than 5min... And where was the class nerd to anser this one?
Also, like I said before she doesn't do just math, she teaches all our subjects.
markee: using yours, if n=4 then p=21. It should = 23.
RE: Cant get this math. by andrey on 04-27-2008 at 12:43 PM
quote:eh ?
Originally posted by markee
p = (n+2)(n/2+1.5)
much easier:
n(n+1)
2
for n=1,2,3...
The sequence begins with 1, i.e. 1, 3, 6, 10, 15, 21, 28 etc.
(Triangular numbers btw.)
RE: Cant get this math. by markee on 04-27-2008 at 01:10 PM
quote:I think you should try that again
Originally posted by Bilbo
quote:Thanks, and just ftw I am the class nerd.
Originally posted by markee
1) p = (n+2)(n/2+1.5)
I realise it isn't the easiest to find, but it shouldn't take a maths teacher more than 5min... And where was the class nerd to anser this one?
Also, like I said before she doesn't do just math, she teaches all our subjects.
markee: using yours, if n=4 then p=21. It should = 23.
RE: Cant get this math. by Baggins on 04-27-2008 at 01:22 PM
markee: oops, if n=4 then p needs to =15.
andrey: no, yours doesn't work. it needs to start at 3.
RE: Cant get this math. by foaly on 04-27-2008 at 01:30 PM
quote:
Originally posted by andrey
quote:eh ?
Originally posted by markee
p = (n+2)(n/2+1.5)
n(n+1)
2
for n=1,2,3...
That works for the first pattern. The sequence begins with 1, i.e. 1, 3, 6, 10, 15, 21, 28 etc.
(Triangular numbers btw.)
ehm neither work for 1=n...
p = (1+2)(.5+1.5) = 3*2 = 6
p = (1(1+1))/2 = 1
and it starts with 3...
the correct one is ((n+1)(n+2))/2
RE: Cant get this math. by Volv on 04-27-2008 at 01:55 PM
For markee's just replace 'n' with 'n-1' and it will work perfectly fine as well.
andrey's/foaly's looks more elegant though
RE: Cant get this math. by John Anderton on 04-27-2008 at 03:45 PM
quote:Wrong.. Close though
Originally posted by andrey
n(n+1)
2
(n+2)(n+3)/2
quote:sequence starts from 8; so for n=0, I'd want 8. Then further, you said it should be multiplied by 2.
Originally posted by Bilbo
q=2^(n+2)
So the answers are..
1) (n+2)*(n+3)/2
2) 2^(n+3)
(sorry for taking this long but my net died
)EDIT: This is assuming that for n=0, you want the first numbers in the sequence and then go hence forth
If you want those values for n=1, the answers would be..
1) (n+1)*(n+2)/2
2) 2^(n+2)
Explanation for answers... (assuming my initial assumption of first value coming at n=0 which is how it should be imo)
1) 1+2+3+...+n = n*(n+1)/2
for n=0, you have value 3 (which is 1+2)
The sequence hence forth remains the same. So basically for n=0, the sequence will give you the same output as n=2 in the above equation.. so just shift your equation by 2 places!
Hence the answer is (n+2)*(n+3)/2
2) 8 = 2^3
Hence forth, values are multiples of 2.
So for n=0, you want 2^3. n=1, you want 2^4 ie 2^(n+3). So on and so forth.
The answer is 2^(n+3)
RE: Cant get this math. by foaly on 04-27-2008 at 09:20 PM
quote:
Originally posted by John Anderton
quote:Wrong.. Close though
Originally posted by andrey
n(n+1)
2
(n+2)(n+3)/2
edit: I saw wrong...
you copied markees answer...
RE: Cant get this math. by markee on 04-28-2008 at 01:07 PM
quote:All sequences start at zero, even JA knows that...
Originally posted by foaly
quote:
Originally posted by andrey
quote:eh ?
Originally posted by markee
p = (n+2)(n/2+1.5)
n(n+1)
2
for n=1,2,3...
That works for the first pattern. The sequence begins with 1, i.e. 1, 3, 6, 10, 15, 21, 28 etc.
(Triangular numbers btw.)
ehm neither work for 1=n...
p = (1+2)(.5+1.5) = 3*2 = 6
p = (1(1+1))/2 = 1
and it starts with 3...
the correct one is ((n+1)(n+2))/2
RE: Cant get this math. by foaly on 04-28-2008 at 01:21 PM
quote:the ones in the example didn't
Originally posted by markee
quote:All sequences start at zero, even JA knows that...
Originally posted by foaly
quote:
Originally posted by andrey
quote:eh ?
Originally posted by markee
p = (n+2)(n/2+1.5)
n(n+1)
2
for n=1,2,3...
That works for the first pattern. The sequence begins with 1, i.e. 1, 3, 6, 10, 15, 21, 28 etc.
(Triangular numbers btw.)
ehm neither work for 1=n...
p = (1+2)(.5+1.5) = 3*2 = 6
p = (1(1+1))/2 = 1
and it starts with 3...
the correct one is ((n+1)(n+2))/2
RE: Cant get this math. by John Anderton on 04-28-2008 at 05:15 PM
quote:They start with n=0, fool
Originally posted by foaly
quote:the ones in the example didn't
Originally posted by markee
quote:All sequences start at zero, even JA knows that...
Originally posted by foaly
quote:
Originally posted by andrey
quote:eh ?
Originally posted by markee
p = (n+2)(n/2+1.5)
n(n+1)
2
for n=1,2,3...
That works for the first pattern. The sequence begins with 1, i.e. 1, 3, 6, 10, 15, 21, 28 etc.
(Triangular numbers btw.)
ehm neither work for 1=n...
p = (1+2)(.5+1.5) = 3*2 = 6
p = (1(1+1))/2 = 1
and it starts with 3...
the correct one is ((n+1)(n+2))/2
He said, initial value is 3. Implies for n=0, value=3; Then next value is 6 implies for n=1, value=6.
markee, the JA knows *
RE: Cant get this math. by Svip on 04-28-2008 at 05:20 PM
E.g.
code:
<?php
$array = array(3, 6, 9, 12);
foreach($array as $k => $v)
echo "$k => $v;\n"
?>
Output
code:
0 => 3;
1 => 6;
2 => 9;
3 => 12;
Learn to array...
RE: Cant get this math. by John Anderton on 04-28-2008 at 05:35 PM
Programing languages (and the Svip, more importantly ) agree.
Our hypothesis of zeroness must be correct! 
RE: Cant get this math. by somelauw on 04-28-2008 at 05:43 PM
I solved them too by taking the formulas I know and then adjusting them to the ones similair by your sequences:
code:
1,3,6,10
y = (number/2) * (first + last)
y = (x/2)(x+1) = (x*x +x)/2
3, 6, 10, 15, 21
y = (x+1)(x+2)/2 = (x*x + 3x + 2)/2
1,2,4,8
y = g ** x
y = 2 ** (x - 1)
8, 16, 32, 64, 128
y = 2 ** (x + 2)
Edit: I assumed y(1) is the first value in both sequences.
Edit2: Power/multiplication typo. I prefer ** for powers.
RE: Cant get this math. by John Anderton on 04-28-2008 at 05:57 PM
quote:
Originally posted by somelauw
I solved them too by taking the formulas I know and then adjusting them to the ones similair by your sequences:
code:
1,3,6,10
y = (number/2) * (first + last)
y = (x/2)(x+1) = (x*x +x)/2
3, 6, 10, 15, 21
y = (x+1)(x+2)/2 = (x*x + 3x + 2)/2
1,2,4,8
y = g * x
y = 2 * (x - 1)
8, 16, 32, 64, 128
y = 2 * (x + 2)
Firstly, refer to my post earlier in this thread.
Secondly, expression 1 is wrong. I mean the pattern you've considered isn't the one he wants...
The first expression is correct since its a standard. Its usually written as x*(x+1)/2
In the second expression, you are missing the condition where the first value (initial value) is to be taken as x=0; Your expression would be correct if you would have considered that.
(read my post for a complete explanation)Third expression is horribly wrong. You've mixed up multiplication and exponentiation
Fourth expression is wrong because the series is the power of 2 series. You're multiplying 2 with numbers like 3,4,5,6 which won't give you the answers. What you needed was the raised to (also called exponential; noted as ^) and not the multiplication sign. If you replace the multiplication by exponentiation and consider the zero criteria (as mentioned in my post, your answer would be correct)
RE: Cant get this math. by somelauw on 04-28-2008 at 06:24 PM
JA, you're right. I corrected my mistakes.
I didn't remember whether 0 or 1 should be taken as first number for x. I though that both where acceptable.
For powering I prefer the ** sign, because ^ could also stand for bitwise xor. I only made a slight typo by forgetting to double the *.
RE: Cant get this math. by John Anderton on 04-28-2008 at 06:26 PM
quote:yeah, ^ is bitwise ex-or but that's in programming, something we're not referring to now. ** would have worked too
Originally posted by somelauw
For powering I prefer the ** sign, because ^ could also stand for bitwise xor. I only made a slight typo by forgetting to double the *.
RE: RE: Cant get this math. by Pr0xY on 04-28-2008 at 07:21 PM
quote:Well it is Canada.
Originally posted by Phillip
quote:Wow your math teacher must be kinda stupid =\
Originally posted by Bilbo
There are two patterns that I couldn't figure out, and nither could my teacher when I asked her.
.. J/k!!
RE: Cant get this math. by foaly on 04-28-2008 at 08:16 PM
quote:
Originally posted by John Anderton
They start with n=0, fool (Smilie)
He said, initial value is 3. Implies for n=0, value=3; Then next value is 6 implies for n=1, value=6.
quote:1*5+3 =8
Originally posted by Bilbo
Pattern: 8, 13, 18, 23, 18, 33
Description: Start at 8 then add 5.
Expression: n5+3
and my name is foaly not fool
RE: Cant get this math. by markee on 04-28-2008 at 08:51 PM
quote:Well except if you deal with maths programs like matlab, then it starts from 1, but that is becausethey store every variable in a matrix (2 dimensional array)
Originally posted by John Anderton
Programing languages (and the Svip, more importantly
Our hypothesis of zeroness must be correct!