Decrease the space between formula and text, and the formula position of align environment

Question

Please see the picture first. My two questions are described in the picture.

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} 

Answer 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.

\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

Answer 2

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}