1

Please see the picture first. My two questions are described in the picture. enter image description here

  1. As known to us, when we use "\begin{align}", it will appear at the next line automatically. Could I just move it just be after the "(2)"?

  2. About the decreasing the space. I have tried to use "\begin{equation}", see http://tex.stackexchange.com/questions/132965/different-space-between-align-and-equation. However, I fail to compile("there is no line here to end") if there're more than 1 equation in this case. Shall we have other method to make it?

My code:

\documentclass[CJK]{beamer} 
\usepackage{beamerthemesplit} 
\usepackage{CJK,CJKnumb}
\usepackage[english]{babel}
\usepackage{amsmath}
\begin{document}
\begin{CJK*}{GBK}{song} %some words are Chinese.
\begin{frame}
\frametitle{**********************************************************************
(1)Question;\\
(2)Question;\\
(3)Prove.
 }
(1)$2S_1=a_2-\frac{1}{3}-1-\frac{2}{3}=a_2-2$,$S_1=a_1=1 \Rightarrow a_2=4$.\\
(2)
\begin{align}
2S_n&=na_{n+1}-\frac{1}{3}n^3-n^2-\frac{2}{3}n,~~~(n \ge 2) \\[-1.5ex] 
\Rightarrow 2S_{n-1}&=(n-1)a_n-\frac{1}{3}(n-1)^3-(n-1)^2-\frac{2}{3}(n-1)
\end{align}
(1)-(2)得$2a_n=na_{n+1}-(n-1)a_n-\frac{1}{3}(3n^2-3n+1)-(2n-1)-\frac{2}{3}$,\\
整理$(n+1)a_n=na_{n+1}-n(n+1)$,即$\frac{a_{n+1}}{n+1}-\frac{a_n}{n}=1$,又$\frac{a_2} {2}-\frac{a_1}{1}=1$,\\
∴数列$\{\frac{a_n}{n}\}$是以首项$\frac{a_1}{1}$为1,公差为1的等差数列,\\
所以$\frac{a_n}{n}=1+(n-1)\times 1=n \Rightarrow a_n=n^2$.
\end{frame}
\end{CJK*}
\end{document} 
  • Put the equation inside a \parbox{\textwidth} and put that inside a \raisebox. – John Kormylo Nov 19 '13 at 4:20
1

This is an alternative to achieve the goal. (Compiled with XeLaTeX without Chinese shown)

  1. Put the first equation into the align environment too.
  2. Use \phantom{} technique for alignments.
  3. use \vskip -\abovedisplayskip to move the content up.

enter image description here

\usepackage{beamerthemesplit} 
\usepackage{CJK,CJKnumb}
\usepackage[english]{babel}
\usepackage{amsmath}
\begin{document}
\begin{CJK*}{GBK}{song} %some words are Chinese.
\begin{frame}
\frametitle{**********************************************************************
(1)Question;\\
(2)Question;\\
(3)Prove.}
\vskip -4\abovedisplayskip   %%%% newly added
\begin{align}
(1)\phantom{\Rightarrow}2S_1 &=a_2-\frac{1}{3}-1-\frac{2}{3}=a_2-2,S_1=a_1=1 \Rightarrow a_2=4. \nonumber\\
(2)\phantom{\Rightarrow}2S_n & =na_{n+1}-\frac{1}{3}n^3-n^2-\frac{2}{3}n,~~~(n \ge 2) \\[-1.5ex] 
\Rightarrow 2S_{n-1}         &=(n-1)a_n-\frac{1}{3}(n-1)^3-(n-1)^2-\frac{2}{3}(n-1)
\end{align}    
\vskip -\abovedisplayskip    %%%% newly added
(1)-(2)得$2a_n=na_{n+1}-(n-1)a_n-\frac{1}{3}(3n^2-3n+1)-(2n-1)-\frac{2}{3}$,\\
整理$(n+1)a_n=na_{n+1}-n(n+1)$,即$\frac{a_{n+1}}{n+1}-\frac{a_n}{n}=1$,又$\frac{a_2} {2}-\frac{a_1}{1}=1$,\\
∴数列$\{\frac{a_n}{n}\}$是以首项$\frac{a_1}{1}$为1,公差为1的等差数列,\\
所以$\frac{a_n}{n}=1+(n-1)\times 1=n \Rightarrow a_n=n^2$.
\end{frame}
\end{CJK*}
\end{document} 

Edit: At last, got XeLaTeX working for Chinese and this is how it looks

enter image description here

1

I used xelatex instead of pdflatex which makes live easier with CJK:

\documentclass[fleqn,10pt]{beamer} 
\usepackage{beamerthemesplit} 
\usepackage{amsmath}
\usepackage{fontspec}
\setsansfont{Code2000}
\usepackage[english]{babel}
\begin{document}

\begin{frame}
\frametitle{**********************************************************************
(1)Question;\\
(2)Question;\\
(3)Prove.}
%
\vspace*{-5ex}\belowdisplayskip=0pt
\begin{flalign}
(1) && 2S_1 &=a_2-\frac{1}{3}-1-\frac{2}{3}=a_2-2, \,S_1=a_1=1 \Rightarrow a_2=4.\nonumber\\
(2) && \,2S_n&=na_{n+1}-\frac{1}{3}n^3-n^2-\frac{2}{3}n,~~~(n \ge 2) \\
    &&      \Rightarrow 2S_{n-1}&=(n-1)a_n-\frac{1}{3}(n-1)^3-(n-1)^2-\frac{2}{3}(n-1)
\end{flalign}
(1)-(2)得$2a_n=na_{n+1}-(n-1)a_n-\frac{1}{3}(3n^2-3n+1)-(2n-1)-\frac{2}{3}$,\\[1ex]
整理$(n+1)a_n=na_{n+1}-n(n+1)$,即$\frac{a_{n+1}}{n+1}-\frac{a_n}{n}=1$,又$\frac{a_2} {2}-\frac{a_1}{1}=1$,\\[1ex]
∴数列$\{\frac{a_n}{n}\}$是以首项$\frac{a_1}{1}$为1,公差为1的等差数列,\\[1ex]
所以$\frac{a_n}{n}=1+(n-1)\times 1=n \Rightarrow a_n=n^2$.

\end{frame}

\end{document} 

enter image description here

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