Shoutbox

Hehe programmers are usually good with math...

quote:

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Originally posted by Daryl
Hi may73alliance

The formula to calculate term n is (n^2) * 3.

Hope ur gonna have a good grade on this

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lol so do i don't look like it though, lol owel

Well Daryl's answer is right but I got the B section differently (although a little more confusing.)

My result was:
           n= [(n-1)-(n-2)+6] + (n-1)

eg.
In this example I will use 108 as n, therefore (n-1) = 75 and (n-2) is 48
We want to calculate n so i will write the equation as
             n= (75-48+6)+75
therefore n=108

I know my method takes more effort to reach the same answer that Daryl got but depending on the wording of the question, you may not be allowed to put the nth term on each side as is done in Daryls...
Hope this helps

quote:

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Originally posted by Dempsy
well its n - (n-1)=6

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It's actually (n-1)-(n-2)=6

quote:

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Originally posted by inc_haydn
Hope this helps

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its helped me undertsand it but shame cant use it on the bloody exam. lol, i think that is kinda hard for the 1st question though

quote:

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Originally posted by inc_haydn

quote:

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Originally posted by Dempsy
well its n - (n-1)=6

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It's actually (n-1)-(n-2)=6

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well i was close lol

quote:

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Originally posted by inc_haydn
It's actually (n-1)-(n-2)=6

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that formula doesn't work either, remember it's a sequence, not a normal algebratic expression, so n doesn't equal any random number, n equals the nth number. so...

0, 1  , 2  , 3  , 4  , 5  , 6
3, 12, 27, 48, 75, __, __.

for 3, n=0, for 75, n=4 and so on.

I can't solve it right now, cause i've got school, but have fun

See the sequence is this ....
3, 12, 27, 48, 75, 108, 147

(1^2)*3 = 1 * 3 = 3
(2^2)*3 = 4 * 3 = 12
(3^2)*3 = 9 * 3 = 27
(4^2)*3 = 16 * 3 = 48
(5^2)*3 = 25 * 3 = 75
(6^2)*3 = 36 * 3 = 108
(7^2)*3 = 49 * 3 = 147

Nth term is (n^2)*3

ie ... n square into 3 (for ppl who may not understand the symbols but im sure most ppl do)

quote:

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Originally posted by John Anderton
See the sequence is this ....
3, 12, 27, 48, 75, 108, 147

(1^2)*3 = 1 * 3 = 3
(2^2)*3 = 4 * 3 = 12
(3^2)*3 = 9 * 3 = 27
(4^2)*3 = 16 * 3 = 48
(5^2)*3 = 25 * 3 = 75
(6^2)*3 = 36 * 3 = 108
(7^2)*3 = 49 * 3 = 147

Nth term is (n^2)*3

ie ... n square into 3 (for ppl who may not understand the symbols but im sure most ppl do)

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Sequences start with 0 though, you're close, but the 0 need to equal 3

lol, this is beyond a joke. now i've looked at it how JA has set it out its soooo bloody simple......is there a emoticon with a dunce hat? *place one here*

EDIT ok now Ddunk sayin JA's is wrong, ohhh sooooo confused !

hehe just use the green formula I put and ignore what i wrote correcting Dempsy...

          n= [(n-1)-(n-2)+6] + (n-1)

And this does work. As for DDunk saying 0 is the first number that also works with this equation.

If I used the 1st figure (3) and substitute all of the blue numbers below into my equation, this is what results:

n=3
(n-1)=0
(n-2)=3 (I got this answer because it must be the opposite of the 3 ie the system is a quadratic)

      therefore n= (0-3+6)+0
                        =3

**********
 
Actual value of n          -2         -1         0           1           2        3             4              5

Actual number value      12         3          0            3         12       27          48           75
                                 
No.s used in example   (n-3)     (n-2)    (n-1)        (n)        (n+1)      (n+2)      (n+3)     (n+4)