What happened to the Messenger Plus! forums on msghelp.net?
Shoutbox » MsgHelp Archive » General » General Chit Chat » Cant get this math.

Pages: (3): « First « 1 [ 2 ] 3 » Last »
Cant get this math.
Author: Message:
markee
Veteran Member
*****

Avatar

Posts: 1621
Reputation: 50
35 / Male / Flag
Joined: Jan 2006
RE: Cant get this math.
quote:
Originally posted by Bilbo
quote:
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?
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.
I think you should try that again ;)
[Image: markee.png]
04-27-2008 01:10 PM
Profile PM Find Quote Report
Baggins
Full Member
***

Avatar
B000ALFAZO

Posts: 387
Reputation: 13
29 / Male / Flag
Joined: Oct 2006
O.P. RE: Cant get this math.
markee: oops, if n=4 then p needs to =15.
andrey: no, yours doesn't work. it needs to start at 3.
04-27-2008 01:22 PM
Profile E-Mail PM Web Find Quote Report
foaly
Senior Member
****

Avatar

Posts: 718
Reputation: 20
37 / Male / Flag
Joined: Jul 2006
RE: Cant get this math.
quote:
Originally posted by andrey
quote:
Originally posted by markee
p = (n+2)(n/2+1.5)
eh ? :p



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
04-27-2008 01:30 PM
Profile E-Mail PM Find Quote Report
Volv
Skinning Contest Winner
*****

Avatar

Posts: 1233
Reputation: 31
34 / Male / Flag
Joined: Oct 2004
RE: Cant get this math.
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 :)

This post was edited on 04-27-2008 at 01:56 PM by Volv.
04-27-2008 01:55 PM
Profile PM Find Quote Report
John Anderton
Elite Member
*****

Avatar

Posts: 3908
Reputation: 80
36 / Male / Flag
Joined: Nov 2004
Status: Away
RE: Cant get this math.
quote:
Originally posted by andrey
n(n+1)
   2
Wrong.. Close though
(n+2)(n+3)/2 :P

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

This post was edited on 04-27-2008 at 03:52 PM by John Anderton.
[

KarunAB.com
]

[img]http://gamercards.exophase.com/459422.png[
/img]
04-27-2008 03:45 PM
Profile E-Mail PM Web Find Quote Report
foaly
Senior Member
****

Avatar

Posts: 718
Reputation: 20
37 / Male / Flag
Joined: Jul 2006
RE: Cant get this math.
quote:
Originally posted by John Anderton
quote:
Originally posted by andrey
n(n+1)
   2
Wrong.. Close though
(n+2)(n+3)/2 :P


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


This post was edited on 04-28-2008 at 01:20 PM by foaly.
04-27-2008 09:20 PM
Profile E-Mail PM Find Quote Report
markee
Veteran Member
*****

Avatar

Posts: 1621
Reputation: 50
35 / Male / Flag
Joined: Jan 2006
RE: Cant get this math.
quote:
Originally posted by foaly
quote:
Originally posted by andrey
quote:
Originally posted by markee
p = (n+2)(n/2+1.5)
eh ? :p



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
All sequences start at zero, even JA knows that... :dodgy:
[Image: markee.png]
04-28-2008 01:07 PM
Profile PM Find Quote Report
foaly
Senior Member
****

Avatar

Posts: 718
Reputation: 20
37 / Male / Flag
Joined: Jul 2006
RE: Cant get this math.
quote:
Originally posted by markee
quote:
Originally posted by foaly
quote:
Originally posted by andrey
quote:
Originally posted by markee
p = (n+2)(n/2+1.5)
eh ? :p



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
All sequences start at zero, even JA knows that... :dodgy:
the ones in the example didn't :P
04-28-2008 01:21 PM
Profile E-Mail PM Find Quote Report
John Anderton
Elite Member
*****

Avatar

Posts: 3908
Reputation: 80
36 / Male / Flag
Joined: Nov 2004
Status: Away
RE: Cant get this math.
quote:
Originally posted by foaly
quote:
Originally posted by markee
quote:
Originally posted by foaly
quote:
Originally posted by andrey
quote:
Originally posted by markee
p = (n+2)(n/2+1.5)
eh ? :p



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
All sequences start at zero, even JA knows that... :dodgy:
the ones in the example didn't :P
They start with n=0, fool :P
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 * :P
[

KarunAB.com
]

[img]http://gamercards.exophase.com/459422.png[
/img]
04-28-2008 05:15 PM
Profile E-Mail PM Web Find Quote Report
Svip
Board's Foremost Baby-Eater
***

Avatar
I eat babies.

Posts: 101
Reputation: 29
37 / Other / Flag
Joined: Jul 2005
RE: Cant get this math.
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...
[Image: weredoomed4ds.png]
04-28-2008 05:20 PM
Profile E-Mail PM Web Find Quote Report
Pages: (3): « First « 1 [ 2 ] 3 » Last »
« Next Oldest Return to Top Next Newest »


Threaded Mode | Linear Mode
View a Printable Version
Send this Thread to a Friend
Subscribe | Add to Favorites
Rate This Thread:

Forum Jump:

Forum Rules:
You cannot post new threads
You cannot post replies
You cannot post attachments
You can edit your posts
HTML is Off
myCode is On
Smilies are On
[img] Code is On