Shoutbox

quote:

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Originally posted by Bilbo

quote:

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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?

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Thanks, and just ftw I am the class nerd.

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.

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I think you should try that again

markee: oops, if n=4 then p needs to =15.
andrey: no, yours doesn't work. it needs to start at 3.

quote:

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Originally posted by andrey

quote:

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Originally posted by markee
p = (n+2)(n/2+1.5)

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eh ?



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.)

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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

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

quote:

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Originally posted by andrey
n(n+1)
   2

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Wrong.. Close though
(n+2)(n+3)/2

quote:

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Originally posted by Bilbo
q=2^(n+2)

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sequence starts from 8; so for n=0, I'd want 8. Then further, you said it should be multiplied by 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)

quote:

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Originally posted by John Anderton

quote:

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Originally posted by andrey
n(n+1)
   2

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Wrong.. Close though
(n+2)(n+3)/2

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edit: I saw wrong...
you copied markees answer...

quote:

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Originally posted by foaly

quote:

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Originally posted by andrey

quote:

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Originally posted by markee
p = (n+2)(n/2+1.5)

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eh ?



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.)

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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

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All sequences start at zero, even JA knows that...

quote:

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Originally posted by markee

quote:

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Originally posted by foaly

quote:

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Originally posted by andrey

quote:

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Originally posted by markee
p = (n+2)(n/2+1.5)

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eh ?



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.)

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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

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All sequences start at zero, even JA knows that...

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the ones in the example didn't

quote:

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Originally posted by foaly

quote:

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Originally posted by markee

quote:

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Originally posted by foaly

quote:

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Originally posted by andrey

quote:

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Originally posted by markee
p = (n+2)(n/2+1.5)

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eh ?



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.)

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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

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All sequences start at zero, even JA knows that...

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the ones in the example didn't

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They start with n=0, fool
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 *

E.g.

code:

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<?php
$array = array(3, 6, 9, 12);

foreach($array as $k => $v)
  echo "$k => $v;\n"
?>

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Output

code:

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0 => 3;
1 => 6;
2 => 9;
3 => 12;

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Learn to array...