Question

How can the next term in the infinite sequence 1, 5, 12, 22, 35, be generated? O Square the term number, subtract the term number from the result, multiply by 3, and divide the result by 2. O Square the term number, multiply the result by 3, divide by 2, and subtract the term number from the result. O Square the term number, divide the result by 2, subtract the term number, and multiply the result by 3. O Square the term number, multiply the result by 3, subtract the term number, and divide the result by 2.

Answer 1

Check the forward differences of the sequence.

• first-order differences

5 - 1 = 4

12 - 5 = 7

22 - 12 = 10

35 - 22 = 13

• second-order differences (i.e. differences of the first differences)

7 - 4 = 3

10 - 7 = 3

13 - 10 = 3

The second differences are all 3 (as far as we know), so the sequence of first differences is arithmetic/linear, which means the original sequence is quadratic. Let the $n$-th term be

$x_n = an^2 + bn + c$

Given that $x_1=1$, $x_2=5$, and $x_3=12$, we have

$\begin{cases} a + b + c = 1 \\ 4a + 2b + c = 5 \\ 9a + 3b + c = 12 \end{cases} \implies a=\dfrac32, b=-\dfrac12, c=0$

and so the $n$-th term of the sequence is generated by the rule

$x_n = \dfrac{3n^2 - n}2$

which most closely resembles the last option,

Square the term number, multiply the result by 3, subtract the term number, and divide the result by 2.