Need help finding next row in number pattern

jjs_mcg

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Sep 19, 2012
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consider the following table
022200000
024642000
026121412620
02820323832208
?????????
find a pattern and predict the next row of the table

FYI-no one in our grade 12 academic course has been able to solve it please help me!
 
Noone? What a bleak future we have. :p Pascal might inspire you to the answer.
 
Hello, jjs_mcg!

Some of the "patterns" are non-existent.


Consider the following table.

. . 12345678902220000002464200002612141262002820323832208?????????\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline _1&_2&_3&_4&_5&_6&_7&_8&_9 \\ \hline \hline 0&2&2&2&0&0&0&0&0 \\ \hline 0&2&4&6&4&2&0&0&0 \\ \hline 0&2&6&12&14&12&6&2&0 \\ \hline 0&2&8&20&32&38&32&20&8\\ \hline ?&?&?&?&?&?&?&?&? \\ \hline \end{array}

Find a pattern and predict the next row of the table

FYI - no one in our grade 12 academic course has been able to solve it. Please help me!

In column-1, I assume the next number is 0.\displaystyle 0.

Similarly, the next number in column-2 is 2.\displaystyle 2.


Column-3 has: 2, 4, 6, 8, ...
The numbers "go up by 2".
So the next number is 10.\displaystyle 10.


Column-4 has: 2, 6, 12, 20, ...
Consider the difference of consecutive terms

Sequence261220Difference468\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 10\displaystyle 10 and the next number is 30.\displaystyle 30.


Column-5 has: 0, 4, 14, 32, ...

Sequence0414321st diff410182nd diff68\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 10\displaystyle 10, the next 1st difference is 28.\displaystyle 28.
. . Hence, the next number is 60.\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 f(n)\displaystyle f(n) so that: .{f(1)=0f(2)=0f(3)=0f(4)=8}\displaystyle \begin{Bmatrix}f(1) \:=\:0 \\ f(2) \:=\:0 \\ f(3) \:=\:0 \\ f(4) \:=\:8 \end{Bmatrix}

One such function is: .f(n)=43(n36n2+11n6)\displaystyle f(n) \:=\:\tfrac{4}{3}\left(n^3 - 6n^2 + 11n - 6\right)

Therefore: .f(5)=43(125150+556)=43(24)=32\displaystyle f(5) \:=\:\tfrac{4}{3}(125 - 150 + 55 - 6) \:=\:\tfrac{4}{3}(24) \:=\:32
 

Some of the "patterns" are non-existent. False.
Therefore: .f(5)=43(125150+556)=43(24)=32\displaystyle f(5) \:=\:\tfrac{4}{3}(125 - 150 + 55 - 6) \:=\:\tfrac{4}{3}(24) \:=\:32
Waaaay too complicated as well as producing the wrong answer.


The solution is straightforward once you see it. Hint: Pascal.
 
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 :(
 
ps just to clarify this is not grade 11 advance functions, its a mixed credit for student who aren't so savy around mathematics like myself
 
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?
 
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