Ryan has playing cards which he stacks into pyramids. He will begin by stacking up cards to form a pyramid with 1 level, followed by another pyramid with 2 levels and so forth. The pyramids with n levels for different values of n are shown below: Let SnS_n denotes the number of cards in a pyramid with nn levels. It is given that Sn=an2+bn+cS_n = an^2 + bn + c for some constants a,ba,b and $c.

  1. (a)

    Give an expression of the number of additional cards needed to form a pyramid of nnth level from (n−1)(n-1)th level. Leave your expression in terms of aa, bb and nn.

  2. (ii)

    Find the values of aa, bb and cc.

  3. (iii)

    Hence prove that SnS_n is the sum of an arithmetic progression and state the common difference.

  4. (iv)

    One pyramid of each level from 1 to 23 is formed. Find the total number of cards required to form these 23 pyramids.

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Worked solution

1(a)

1. Identify that the number of additional cards needed is the difference between the total cards in the $n$th pyramid and the $(n-1)$th pyramid.

Additional cards=Sn−Sn−1\text{Additional cards} = S_n - S_{n-1}

2. Substitute the algebraic expressions for $S_n$ and $S_{n-1}$ and simplify.

Sn−Sn−1=(an2+bn+c)−[a(n−1)2+b(n−1)+c]=an2+bn+c−[a(n2−2n+1)+bn−b+c]=an2+bn+c−an2+2an−a−bn+b−c=2an−a+b\begin{align*} S_n - S_{n-1} &= (an^2 + bn + c) - [a(n-1)^2 + b(n-1) + c] \\ &= an^2 + bn + c - [a(n^2 - 2n + 1) + bn - b + c] \\ &= an^2 + bn + c - an^2 + 2an - a - bn + b - c \\ &= 2an - a + b \end{align*}
Answer

2an−a+b2an - a + b

1(ii)

1. Determine the number of cards $S_1$, $S_2$, and $S_3$ by logically counting the cards from the described pyramids.

By counting the cards from the structural pattern of the pyramids:
Level 1 (n=1n=1): 2 leaning cards.
S1=2S_1 = 2

Level 2 (n=2n=2): The previous pyramid, plus 1 flat base card, plus 4 leaning cards at the bottom.
S2=2+1+4=7S_2 = 2 + 1 + 4 = 7

Level 3 (n=3n=3): The previous pyramid, plus 2 flat base cards, plus 6 leaning cards at the bottom.
S3=7+2+6=15S_3 = 7 + 2 + 6 = 15

2. Formulate a system of linear equations by substituting $n=1, 2, 3$ into the given quadratic expression.

Substitute n=1,2,3n=1, 2, 3 into the given equation Sn=an2+bn+cS_n = an^2 + bn + c:

For n=1n=1:

a(1)2+b(1)+c=2  ⟹  a+b+c=2— (1)a(1)^2 + b(1) + c = 2 \implies a + b + c = 2 \quad \text{--- (1)}

For n=2n=2:

a(2)2+b(2)+c=7  ⟹  4a+2b+c=7— (2)a(2)^2 + b(2) + c = 7 \implies 4a + 2b + c = 7 \quad \text{--- (2)}

For n=3n=3:

a(3)2+b(3)+c=15  ⟹  9a+3b+c=15— (3)a(3)^2 + b(3) + c = 15 \implies 9a + 3b + c = 15 \quad \text{--- (3)}

3. Solve the system of equations to find the values of $a$, $b$, and $c$.

Subtract (1) from (2) to eliminate cc:
(4a+2b+c)−(a+b+c)=7−2(4a + 2b + c) - (a + b + c) = 7 - 2
3a+b=5— (4)3a + b = 5 \quad \text{--- (4)}

Subtract (2) from (3) to eliminate cc:
(9a+3b+c)−(4a+2b+c)=15−7(9a + 3b + c) - (4a + 2b + c) = 15 - 7
5a+b=8— (5)5a + b = 8 \quad \text{--- (5)}

Subtract (4) from (5) to eliminate bb:
(5a+b)−(3a+b)=8−5(5a + b) - (3a + b) = 8 - 5
2a=3  ⟹  a=1.52a = 3 \implies a = 1.5

Substitute a=1.5a = 1.5 into (4):
3(1.5)+b=53(1.5) + b = 5
4.5+b=5  ⟹  b=0.54.5 + b = 5 \implies b = 0.5

Substitute a=1.5a = 1.5 and b=0.5b = 0.5 into (1):
1.5+0.5+c=21.5 + 0.5 + c = 2
2+c=2  ⟹  c=02 + c = 2 \implies c = 0

Answer

a=1.5,b=0.5,c=0a = 1.5, b = 0.5, c = 0

1(iii)

1. Identify the general term (cards in the nth level) from part (a)

From part (a), the number of additional cards needed to go from the (n−1)(n-1)th to the nnth level is the number of cards in the nnth level:
un=Sn−Sn−1=2an−a+b.u_n = S_n - S_{n-1} = 2an - a + b.
Thus Sn=u1+u2+⋯+unS_n = u_1 + u_2 + \cdots + u_n, i.e. SnS_n is a sum of the terms unu_n.

2. Show consecutive terms have a constant difference and state it

Consider the difference between consecutive terms:
un−un−1=(2an−a+b)−(2a(n−1)−a+b)=2a.u_n - u_{n-1} = (2an - a + b) - \big(2a(n-1) - a + b\big) = 2a.
Since this difference is constant for all nn, the terms unu_n form an arithmetic progression, and therefore Sn=u1+u2+⋯+unS_n = u_1 + u_2 + \cdots + u_n is the sum of an arithmetic progression.

Using a=32a = \tfrac{3}{2} (from part (ii)), the common difference is
2a=2(32)=3.2a = 2\left(\tfrac{3}{2}\right) = 3.

Answer

The number of cards in the nnth level is un=2an−a+bu_n = 2an - a + b, and un−un−1=2au_n - u_{n-1} = 2a (a constant), so the terms form an arithmetic progression. Since Sn=u1+u2+⋯+unS_n = u_1 + u_2 + \cdots + u_n, SnS_n is the sum of this AP. With a=32a = \tfrac{3}{2}, the common difference is 2a=32a = 3.

1(iv)

1. Set up the summation for the total number of cards across the 23 pyramids

Using the sequence expression Sn=1.5n2+0.5nS_n = 1.5n^2 + 0.5n established in the earlier parts, the total number of cards required to build one pyramid of each level from 1 to 23 is the sum of SnS_n:
Total cards=∑n=123Sn=∑n=123(1.5n2+0.5n)\text{Total cards} = \sum_{n=1}^{23} S_n = \sum_{n=1}^{23} (1.5n^2 + 0.5n)

2. Evaluate the summation

Evaluating this finite sum (using a Graphing Calculator or standard formulas for ∑n2\sum n^2 and ∑n\sum n):
Total cards=6624\text{Total cards} = 6624

Answer

6624

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