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=\frac32, b=-\frac12, c=0`

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

`x_n = \frac{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.