MATH SOLVE

6 months ago

Q:
# If $(a_1,a_2,\ldots,a_{17})$ satisfy \begin{align*} a_1 + a_2 + a_3 &= 1, \\ a_2 + a_3 + a_4 &= 2, \\ a_3 + a_4 + a_5 &= 3, \\ &~~\vdots \\ a_{15} + a_{16} + a_{17} &= 15, \\ a_{16} + a_{17} + a_{1} &= 16, \\ a_{17} + a_{1} + a_{2} &= 17, \end{align*} what is the value of $a_{17}$?

Accepted Solution

A:

In this case, your best bet is to use align* for the outer alignment (the = signs), and aligned for the inner one (the + signs): this is because \begin{aligned}...\end{aligned} is designed to be nested within other kinds of display math. However, you also want to vertically align the inner one (at the pluses) with the outer one, so you'll write it as \begin{aligned}[t]...\end{aligned}.
Here's what it looks like in code:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
A_1 &= \begin{aligned}[t]
a_1 &+ a_2 + a_3 + a_4 + a_5 + a_6 + a_7\\
&+ a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13}\\
&+ a_{14} + a_{15} + a_{16} + a_{17} + a_{18} + a_{19}
\end{aligned}\\
A_2 &=
\begin{aligned}[t]
a_2 &+ b_2 + b_3 + b_4 + b_5 + b_6 + b_7\\
&+ b_8 + b_9 + b_{10} + b_{11} + b_{12} + b_{13}\\
&+ b_{14} + b_{15} + b_{16} + b_{17} + b_{18} + b_{19}
\end{aligned}
\end{align*}
\end{document}