Track / Stage 5 / 229 of 400 #829 of 1964

Problem 829

AIME late

Algebra Difficulty 5.4 Find the answer

(15 points) In the sequence $\left\{a_{n}\right\}$ , $a_{n}=2^{n} a+b n-80\left(a, b \in \mathbf{Z}_{+}\right)$ .
It is known that the minimum value of the sum of the first $n$ terms $S_{n}$ is obtained if and only if $n=6$ , and $7 \mid a_{36}$ . Find the value of $\sum_{i=1}^{12}\left|a_{i}\right|$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Three, 13. Notice that, $\left\{a_{n}\right\}$ is an increasing sequence.
From the given, $a_{6} > 0$ , which means
$64 a + 6 b - 800 \text{. }$

Combining $a, b \in \mathbf{Z}_{+}$ , we get
$a=1, b=1$ or 2.
$\begin{array}{l} \text{Also, } a_{36}=2^{36} + 36 b - 80 \\ \equiv 1 + b - 3 \equiv 0 (\bmod 7). \end{array}$

Thus, $b=2$ .
$\begin{array}{l} \text{Therefore, } a_{n}=2^{n} + 2 n - 80 \text{. } \\ \text{Hence, } \left|a_{1}\right| + \left|a_{2}\right| + \cdots + \left|a_{12}\right| \\ = -\left(a_{1} + a_{2} + \cdots + a_{6}\right) + \left(a_{7} + a_{8} + \cdots + a_{12}\right) \\ = S_{12} - 2 S_{6} = 8010. \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.