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# Help

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In stage one, there is 1 block. In stage 2, there are 4 blocks. In stage 3, there are 7 blocks.

()

()                                           ()

()                                     () () ()                                 () () () () ()                             () = Block   s = Stage  n = Blocks

Stage 1                             Stage 2                                    Stage 3

Find a equation other than 3s + 1 = n

Complex  Dec 23, 2015

#4
+92256
+10

Hi Complex,

I plotted the 3 points and decided that a cubic could go through them if there was a horizontal point of inflection at (2,4)

so the graph goes though (1,1). and (3,7) and (2,4)

also     y'(2)=0   and y''(2)=0

so I started with

\(y=ax^3+bx^2+cx+d\\ y'=3ax^2+2bx+c\\ y''=6ax+2b \)

\(0=6a(2) + 2b\\ 0=12a+2b\\ 0=6a+b\\ b=-6a\\~\\ 0=3a(2^2)+2b(2)+c\\ 0=12a+4b+c\\ 0=12a+4(-6a)+c\\ 0=12a-24a+c\\ 0=-12a+c\\ c=12a\\~\\ y=ax^3+bx^2+cx+d\\ y=ax^3-6ax^2+12ax+d\\\)

\(y=ax^3-6ax^2+12ax+d\\ though (1,1)\\ 1=a-6a+12a+d\\ 1=7a+d\\ d=1-7a\\ y=ax^3-6ax^2+12ax+1-7a\\~\\ \)
\(through (2,4)\\ 4=a*2^3-6a*2^2+12a*2+1-7a\\ 4=8a-24a+24a+1-7a\\ 4=a+1\\ a=3\)

\(y=ax^3-6ax^2+12ax+1-7a\\ y=3x^3-18x^2+36x+1-21\\ y=3x^3-18x^2+36x-20\\\)

\(\boxed{n=3s^3-18s^2+36s-20}\)

Melody  Dec 24, 2015
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#1
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IDFGI

Guest Dec 23, 2015
#2
+85958
+10

1  4  7  10  13

1^2  - 0^2 + 0   =  1 + 0   = 1

2^2 - 1^2 + 1   =  3 + 1   = 4

3^2 - 2^2 + 2 =  5 + 2  = 7

4^2 - 3^2 + 3 = 7 + 3  = 10

5^2 - 4^2 + 4 = 9 + 4  = 13

6^2 - 5^2 + 5  =  11 + 5  = 16

=

[ (n)^2 - (n-1)^2] + (n - 1)  =  3n - 2

CPhill  Dec 23, 2015
#3
+92256
0

Chris, I do not think that yours is a new formula, it is just a mor complicated way of writing exactly the same formula.

Melody  Dec 24, 2015
#4
+92256
+10

Hi Complex,

I plotted the 3 points and decided that a cubic could go through them if there was a horizontal point of inflection at (2,4)

so the graph goes though (1,1). and (3,7) and (2,4)

also     y'(2)=0   and y''(2)=0

so I started with

\(y=ax^3+bx^2+cx+d\\ y'=3ax^2+2bx+c\\ y''=6ax+2b \)

\(0=6a(2) + 2b\\ 0=12a+2b\\ 0=6a+b\\ b=-6a\\~\\ 0=3a(2^2)+2b(2)+c\\ 0=12a+4b+c\\ 0=12a+4(-6a)+c\\ 0=12a-24a+c\\ 0=-12a+c\\ c=12a\\~\\ y=ax^3+bx^2+cx+d\\ y=ax^3-6ax^2+12ax+d\\\)

\(y=ax^3-6ax^2+12ax+d\\ though (1,1)\\ 1=a-6a+12a+d\\ 1=7a+d\\ d=1-7a\\ y=ax^3-6ax^2+12ax+1-7a\\~\\ \)
\(through (2,4)\\ 4=a*2^3-6a*2^2+12a*2+1-7a\\ 4=8a-24a+24a+1-7a\\ 4=a+1\\ a=3\)

\(y=ax^3-6ax^2+12ax+1-7a\\ y=3x^3-18x^2+36x+1-21\\ y=3x^3-18x^2+36x-20\\\)

\(\boxed{n=3s^3-18s^2+36s-20}\)

Melody  Dec 24, 2015
#5
+85958
+5

Mmmmmm.....the confusion over "formulas"  arises from the initial value of "s"

Actually, Complex"......if s = the stage number, then your formula  of 3s + 1  is incorrect, because at stage 1, s = 1 and we should have 4  blocks instead of just 1

I made a slight mistake in my formula  by using n instead of s  .......

I should have been....

3s - 2  = n,      assuming we can set s = 1   at the start

But....as to Melody's  assertion that it's the "same" formula re-written, we can see that it's not true because if it were, then

3s + 1  =  3s - 2   which implies that

1 = -2      !!!!

Melody's  "formula"  calculates the first three terms correctly......the 4th term would be 28 and the 5th term would be 85

If the intent of the sequence is to have a difference of 3 blocks at every stage, then her formula would not be correct.......if something else was intended, her formula might be correct and mine might be incorrect....

Who knows???

CPhill  Dec 24, 2015
edited by CPhill  Dec 24, 2015
#6
+92256
+5

Hi Chris and Complex.

I did not check the first formula I just assumed it was correct.

Since the first one was a linear, and Chris's was linear, and if they were both correct they would HAVE to be equivalent.

So I assumed Chris's was the same as the one given. (I did not actually compare them)

Three points defines a line so all linear representions have to be equivalent. ://

Melody  Dec 25, 2015

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