A list of numbers as in Section 1.1 (2, 9, 16, 23, 30) does not mean anything on its own. The context before this example implies that one should look for an arithmetic relationship.
The “trick” is that a premise is withheld. As in poorly written mystery novels, crucial information is not available.
All reasoning is based on premises (hypotheses, suppositions, etc.) wether implicit or explicit. “Trick” questions like Section 1.1’s example rely on misleading you into using an incorrect implicit premise.
3  
14  11  
31  17  6 
54  23  6 
83  29  6 
118  35  6 
159  41  6 
The formula provided in the text is of order 4, or in other words the highest power of the argument n is {n}^{4}. (Another phrase for this is that the formula is quartic.) We expect to need 4 columns to the right of the original sequence (1, 2, 4, 8, 16, 31) to reach an arithmetic sequence.
points  regions  {Δ}^{(1)}  {Δ}^{(2)}  {Δ}^{(3)}  {Δ}^{(4)} 
1  1  
2  2  1  
3  4  2  1  
4  8  4  2  1  
5  16  8  4  2  1 
6  31  15  7  3  1 
7  57  26  11  4  1 
8  99  42  16  5  1 
The formula provided is
R(n) = {1\over
24}\left ({n}^{4}  6{n}^{3} + 23{n}^{2}  18n + 24\right ).

One can compute this directly with any method to verify that the answer is 99.
One convenient way to rewrite a polynomial for evaluation is Horner’s rule. Horner’s rule applies the distributive property of multiplication over addition to pull factors of n out of subexpressions. This rule not only is faster when using a calculator, it also incurs fewer rounding errors when n is not an integer.
Applying Horner’s rule,
Subsituting 8 we find that
Dividing directly again verifies the result is 99, but a technique to avoid the division is recognizing that 2376 = 2400  24. Then
The problem is of order 2 (or is quadratic), so we expect two columns beyond the initial sequence.
n  {n}^{2} + 3n + 1 = (n + 3)n + 1  {Δ}^{(1)}  {Δ}^{(2)} 
1  5  
2  11  6  
3  19  8  2 
4  29  10  2 
5  41  12  2 
Substituting 5 into (n + 3)n + 1 produces (5 + 3) ⋅ 5 + 1 = 8 ⋅ 5 + 1 = 41, verifying the result.
There are two reasonable ways to extend the left pattern. Either is reasonable, and both demonstrate the same property.
The first prepends 10 to each number on the left. The resulting pattern is
The second possiblity “reflects” the number across the leading or trailing 1. The resulting pattern is
The common property is that squaring a number with alternating 1 and 0 digits
With these short sequences, the zeros only serve to make the pattern more obvious. Note that 1{1}^{2} = 121 and 11{1}^{2} = 12\kern 1.66702pt 321. This pattern will break after the central digit is 9. Why?
Note that computing 101010101^{2} with common desktop computers may produce 10203040504030200. The last digit falls off the end of how computers represent floatingpoint numbers. Computing in integers on “32bit” computers may produce 28 if the calculation wraps around the 32bit boundary.
This is one reason why looking for patterns and developing a number sense is important. Errors in calculated results depend on the method used for calculation. Most programs or devices do not explain their methods, so recognizing patterns and other properties are important to prevent being misled.
The next line could well be
One method for verifying the result is simple calculation.
Another is to rearrange the problem slightly to show the pattern
Using the formula {\mathop{\mathop{∑ }}\nolimits }_{i=1}^{n}i = n(n + 1)∕2, the n^{th} middle form is
{n(n + 1)\over
2} + {(n  1)n\over
2} = {{n}^{2} + n + {n}^{2}  n\over
2} = {n}^{2}.

So the fifth term is indeed {5}^{2}.
There are two clear ways to extend the formula S(n) = n(n + 1)∕2 into a formula for the sum 2 + 4 + 6 + \mathrel{⋯} + 2n.
One is to recognize that 2 + 4 + 6 + \mathrel{⋯} + 2n = 2(1 + 2 + 3 + \mathrel{⋯} + n) = 2S(n) = n(n + 1). Rephrasing the original problem using summation notation, we have used {\mathop{\mathop{∑ }}\nolimits }_{i=1}^{n}2i = 2{\mathop{\mathop{∑ }}\nolimits }_{i=1}^{n}i = n(n + 1).
Another is to consider the sum 2+4+6+\mathrel{⋯}+2n = (1+1)+(2+2)+(3+3)+\mathrel{⋯}+(n+n) = (1+2+3+\mathrel{⋯}+n)+(1+2+3+\mathrel{⋯}+n) = S(n)+S(n) = 2S(n) = n(n+1). In summation notation, {\mathop{\mathop{∑ }}\nolimits }_{i=1}^{n}2i ={\mathop{ \mathop{∑ }}\nolimits }_{i=1}^{n}(i + i) = \left ({\mathop{\mathop{∑ }}\nolimits }_{i=1}^{n}i\right ) + \left ({\mathop{\mathop{∑ }}\nolimits }_{i=1}^{n}i\right ) = n(n + 1).
This problem is about phrasing mathematical problems in a way that respects the order of operations.
Some possibilities are the following:
The key is that on the left a sum is squared, while on the right cubes are added.
This sequence of problems demonstrates similar points in different ways. They are all related to each other and to Problem 16.
Each column (or row) of the blocked triangles represents the integers 1, 2, 3, and 4 by the number of blocks in the column (or row). The total number of blocks in each triangle is 1+2+3+4.
When flipped and combined, the total number blocks is the sum of the two triangles, or 2 ⋅ (1 + 2 + 3 + 4). The combined figure is a rectangle consisting of 4 ⋅ 5 blocks. So 2(1 + 2 + 3 + 4) = 4 ⋅ 5, or 1 + 2 + 3 + 4 = (4 ⋅ 5)∕2.
A better diagram would replace the leftmost arrow with an addition operator (+).
Drawing out the dots demonstrates the solution directly.
Or use the result of Problem 16. Note that the first sum, {\mathop{\mathop{∑ }}\nolimits }_{i=1}^{n}i, is the n^{th} triangular number and that the second sum, {\mathop{\mathop{∑ }}\nolimits }_{i=1}^{n1}i, is the (n  1)^{th} triangular number. Thus Problem 16 demonstrated that the sum of two consecutive triangular numbers is a square.
Either draw a few consecutive figures from Problem 39 or use Problem 16.
The first pattern to observe is they are all of the form p(n)∕2, where p(n) is some polynomial of n. The next pattern is that the numerator p(n) = n(a ⋅ n  b) for integers a and b. Then both a and b increase by one when adding a side.
Using these patterns, the n^{th} nonagonal number is
N(n) = {n(7n  5)\over
2} .

Substituting 6, N(6) = 6(7 ⋅ 6  5)∕2 = 3(42  5) = 111, adding more evidence to the conjecture.
Oops. I think meant to give Problem 52 rather than repeat Problem 51, but that’s my fault.
Filling in a few values,
n  T(n  1)  3T(n  1) + n 
2  1  5 
3  3  12 
4  6  22 
5  10  35 
The first, 5, suggests the fivesided pentagon that produces the second pentagonal number. Later numbers add additional evidence.
One could prove the relationship by expanding 3T(n  1) + n and simplifying the expression, or