2008 AMC 10B Problem 11

The rule says each term is twice the term before it plus the term before that. To use it we need two terms in a row, but we know only the 3rd and the 6th. So give the 4th term a name and build forward.

1. Let $u_4 = a$. We know $u_3 = 9$.
2. The rule with $n = 3$: $u_5 = 2u_4 + u_3 = 2a + 9$.
3. The rule with $n = 4$: $u_6 = 2u_5 + u_4 = 2(2a + 9) + a = 5a + 18$.
4. We are told $u_6 = 128$, so
$$5a + 18 = 128, \qquad 5a = 110, \qquad a = 22.$$
5. Then $u_5 = 2(22) + 9 = 53$.
6. Check: $u_6 = 2(53) + 22 = 106 + 22 = 128$. Correct.

The answer is $\boxed{\textbf{(B)}\ 53}$.

A rule that uses the two previous terms needs two terms in a row to get going. When the known terms are spread apart, name one missing term with a letter and write the later terms using it. The rule then turns into an ordinary one-letter equation.

Test the answer choices. If $u_5 = x$, then from $u_6 = 2u_5 + u_4$ we get $u_4 = 128 - 2x$, and the rule also needs $u_5 = 2u_4 + 9$. Try (B) $x = 53$: $u_4 = 128 - 106 = 22$, and $2(22) + 9 = 53$. It fits. (Trying (A) $x = 40$: $u_4 = 48$, but $2(48) + 9 = 105$, not $40$.)

The unknown term between the two given ones is $u_4$; call it $a$. Apply the recurrence twice:
$$u_5 = 2u_4 + u_3 = 2a + 9,$$
$$u_6 = 2u_5 + u_4 = 2(2a + 9) + a = 5a + 18.$$
Setting $u_6 = 128$ gives $5a + 18 = 128$, so $a = 22$. Then
$$u_5 = 2(22) + 9 = 53.$$
Check: $u_6 = 2(53) + 22 = 128$.

The answer is $\boxed{\textbf{(B)}\ 53}$.