+405 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
+118687 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
\(y=ax^3+bx^2+cx+d\\ y'=3ax^2+2bx+c\\ y''=6ax+2b \)
\(0=6a(2) + 2b\\ 0=12a+2b\\ 0=6a+b\\ b=-6a\\~\\ 0=3a(2^2)+2b(2)+c\\ 0=12a+4b+c\\ 0=12a+4(-6a)+c\\ 0=12a-24a+c\\ 0=-12a+c\\ c=12a\\~\\ y=ax^3+bx^2+cx+d\\ y=ax^3-6ax^2+12ax+d\\\)
\(y=ax^3-6ax^2+12ax+d\\ though (1,1)\\ 1=a-6a+12a+d\\ 1=7a+d\\ d=1-7a\\ y=ax^3-6ax^2+12ax+1-7a\\~\\ \)
\(through (2,4)\\ 4=a*2^3-6a*2^2+12a*2+1-7a\\ 4=8a-24a+24a+1-7a\\ 4=a+1\\ a=3\)
\(y=ax^3-6ax^2+12ax+1-7a\\ y=3x^3-18x^2+36x+1-21\\ y=3x^3-18x^2+36x-20\\\)
OR using your letters
\(\boxed{n=3s^3-18s^2+36s-20}\)
+129899 1 4 7 10 13
How about this pattern ????
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
+118687 Chris, I do not think that yours is a new formula, it is just a mor complicated way of writing exactly the same formula. 
+118687 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
\(y=ax^3+bx^2+cx+d\\ y'=3ax^2+2bx+c\\ y''=6ax+2b \)
\(0=6a(2) + 2b\\ 0=12a+2b\\ 0=6a+b\\ b=-6a\\~\\ 0=3a(2^2)+2b(2)+c\\ 0=12a+4b+c\\ 0=12a+4(-6a)+c\\ 0=12a-24a+c\\ 0=-12a+c\\ c=12a\\~\\ y=ax^3+bx^2+cx+d\\ y=ax^3-6ax^2+12ax+d\\\)
\(y=ax^3-6ax^2+12ax+d\\ though (1,1)\\ 1=a-6a+12a+d\\ 1=7a+d\\ d=1-7a\\ y=ax^3-6ax^2+12ax+1-7a\\~\\ \)
\(through (2,4)\\ 4=a*2^3-6a*2^2+12a*2+1-7a\\ 4=8a-24a+24a+1-7a\\ 4=a+1\\ a=3\)
\(y=ax^3-6ax^2+12ax+1-7a\\ y=3x^3-18x^2+36x+1-21\\ y=3x^3-18x^2+36x-20\\\)
OR using your letters
\(\boxed{n=3s^3-18s^2+36s-20}\)
+129899 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???
+118687 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. ://
