Need help finding next row in number pattern
- Thread starter jjs_mcg
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Hello, jjs_mcg!
Some of the "patterns" are non-existent.
In column-1, I assume the next number is \(\displaystyle 0.\)
Similarly, the next number in column-2 is \(\displaystyle 2.\)
Column-3 has: 2, 4, 6, 8, ...
The numbers "go up by 2".
So the next number is \(\displaystyle 10.\)
Column-4 has: 2, 6, 12, 20, ...
Consider the difference of consecutive terms
\(\displaystyle \begin{array}{c|ccccccc} \hline \text{Sequence} & 2 &&6&&12&&20 \\ \hline \text{Difference} &&4&&6&&8 \\ \hline\end{array}\)
I conjecture that the next difference is \(\displaystyle 10\) and the next number is \(\displaystyle 30.\)
Column-5 has: 0, 4, 14, 32, ...
\(\displaystyle \begin{array}{c|ccccccc} \hline \text{Sequence} & 0&&4&&14&&32 \\ \hline \text{1st diff} && 4&&10&&18 \\ \hline \text{2nd diff} &&&6&&8 \\ \hline\end{array}\)
I conjecture that the next 2nd difference is \(\displaystyle 10\), the next 1st difference is \(\displaystyle 28.\)
. . Hence, the next number is \(\displaystyle 60.\)
The rest are too short to find a pattern.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For fun, I applied some algebraic tricks to column-9.
We have: 0, 0, 0, 8 ...
We want a function \(\displaystyle f(n)\) so that: .\(\displaystyle \begin{Bmatrix}f(1) \:=\:0 \\ f(2) \:=\:0 \\ f(3) \:=\:0 \\ f(4) \:=\:8 \end{Bmatrix}\)
One such function is: .\(\displaystyle f(n) \:=\:\tfrac{4}{3}\left(n^3 - 6n^2 + 11n - 6\right) \)
Therefore: .\(\displaystyle f(5) \:=\:\tfrac{4}{3}(125 - 150 + 55 - 6) \:=\:\tfrac{4}{3}(24) \:=\:32\)
Some of the "patterns" are non-existent.
In column-1, I assume the next number is \(\displaystyle 0.\)
Similarly, the next number in column-2 is \(\displaystyle 2.\)
Column-3 has: 2, 4, 6, 8, ...
The numbers "go up by 2".
So the next number is \(\displaystyle 10.\)
Column-4 has: 2, 6, 12, 20, ...
Consider the difference of consecutive terms
\(\displaystyle \begin{array}{c|ccccccc} \hline \text{Sequence} & 2 &&6&&12&&20 \\ \hline \text{Difference} &&4&&6&&8 \\ \hline\end{array}\)
I conjecture that the next difference is \(\displaystyle 10\) and the next number is \(\displaystyle 30.\)
Column-5 has: 0, 4, 14, 32, ...
\(\displaystyle \begin{array}{c|ccccccc} \hline \text{Sequence} & 0&&4&&14&&32 \\ \hline \text{1st diff} && 4&&10&&18 \\ \hline \text{2nd diff} &&&6&&8 \\ \hline\end{array}\)
I conjecture that the next 2nd difference is \(\displaystyle 10\), the next 1st difference is \(\displaystyle 28.\)
. . Hence, the next number is \(\displaystyle 60.\)
The rest are too short to find a pattern.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For fun, I applied some algebraic tricks to column-9.
We have: 0, 0, 0, 8 ...
We want a function \(\displaystyle f(n)\) so that: .\(\displaystyle \begin{Bmatrix}f(1) \:=\:0 \\ f(2) \:=\:0 \\ f(3) \:=\:0 \\ f(4) \:=\:8 \end{Bmatrix}\)
One such function is: .\(\displaystyle f(n) \:=\:\tfrac{4}{3}\left(n^3 - 6n^2 + 11n - 6\right) \)
Therefore: .\(\displaystyle f(5) \:=\:\tfrac{4}{3}(125 - 150 + 55 - 6) \:=\:\tfrac{4}{3}(24) \:=\:32\)
Last edited:
thanks for your help I appreciate it! I still don't see it however! our teacher said once you see it its easy haha. this question is literally killing me I've tried breaking down the difference between values to no avail. I'm so lost I understand pascals triangle vaguely but I don't see the sum of two number in a previous column adding up to a number in the next one 
The symmetry of the rows immediately reminded me of Pascal's triangle. First I tried summing 2 numbers above to get a number below. I got to 4 6 in the 2nd row and there was no 10 in the 3rd row, so throw that plan out the window. There's a 6 in the 2nd row, how do you get that using the first row?