<PRE><B>Question 986170</B>
Draw out a few cases. I drew out pyramids with 1 level, 2 levels, 3 levels, 4 levels, and 5 levels. Basically I am showing below cases n = 1 through n = 5 

&lt;img width=&quot;400px&quot; src = &quot;http://i150.photobucket.com/albums/s91/jim_thompson5910/cards2_zps7mmzrum8.png&quot;&gt;


The list goes on, but eventually we run out of room, so it&#39;s a good place to stop here. We have enough to see a pattern forming.


Notice the sequence of cards needed: 2, 7, 15, 26, 40, ...


The jump from 2 to 7 is +5
The jump from 7 to 15 is +8
The jump from 15 to 26 is +11
The jump from 26 to 40 is +14


We have a list of common differences: 5, 8, 11, 14
Each common difference is separated by 3. 
5+3 = 8
8+3 = 11
11+3 = 14
and so on...


Since we have constant 2nd differences (all 3), this means a quadratic polynomial models the original sequence.


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Pair up this list {1,2,3,4,5} with {2, 7, 15, 26, 40} to get this list of ordered pairs


(1,2), (2,7), (3,15), (4,26), (5,40)


we only need 3 ordered pairs. Let&#39;s just work with the first three ordered pairs. We will work with the general quadratic form y = ax^2+bx+c


Plug in (x,y) = (1,2) into y = ax^2+bx+c to get
y = ax^2+bx+c
2 = a*1^2+b*1+c
a+b+c = 2


Plug in (x,y) = (2,7) into y = ax^2+bx+c to get
y = ax^2+bx+c
7 = a*2^2+b*2+c
4a+2b+c = 7



Plug in (x,y) = (3,15) into y = ax^2+bx+c to get
y = ax^2+bx+c
15 = a*3^2+b*3+c
9a+3b+c = 15


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After plugging in those (x,y) pairs and simplifying, we have this system of equations


a+b+c = 2
4a+2b+c = 7
9a+3b+c = 15


There are a number of ways to solve. I&#39;m going to use a calculator. Specifically the TI83&#39;s inverse matrix feature.


Doing so essentially solves for (a,b,c) simultaneously. The calculator states that


a = 1.5
b = 0.5
c = 0


So the nth term equation is &lt;font size=5 color=&quot;red&quot;&gt;c = 1.5n^2 + 0.5n&lt;/font&gt; where n is the level number and c is the number of cards needed to make the pyramid that tall.


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For instance, if n = 3, then we should get c = 15. Let&#39;s see if that&#39;s true.


c = 1.5n^2 + 0.5n
c = 1.5*3^2 + 0.5*3
c = 1.5*9 + 0.5*3
c = 13.5 + 1.5
c = 15


we definitely get c = 15. So we need 15 cards for a 3 level pyramid. This confirms the equation is true for n = 3. Try other values of n (like n = 1 or n = 2) to generate other values of c. This is a good way to confirm we have the correct nth term formula.
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