\documentclass[11pt,b4j,landscape,twocolumn]{jsarticle}

% jsarticle.cls		http://oku.edu.mie-u.ac.jp/~okumura/jsclasses/ の jsclasses-130514.zip
% b4tate.sty		http://unilab.gbb60166.jp/tex/b4tate.zip
% furikana.sty		http://xymtex.my.coocan.jp/fujitas2/texlatex/index.html#tategumi
%
\usepackage{ascmac,furikana,utf,bm}
\usepackage[dvipdfmx]{graphicx,color}
%\usepackage[dvipdfmx,pdfpagemode=FullScreen,setpagesize=false]{hyperref}
%\usepackage{tikz}
%\usetikzlibrary{intersections, calc, through, quotes, angles}

\textwidth=315mm \textheight=220mm \columnseprule=0pt \topmargin=-25mm \oddsidemargin=-15mm

\newenvironment{nidan}{\begin{minipage}[t]{.225\textwidth}}{\end{minipage}}
\def\nidangumi#1#2{\begin{nidan}{#1}\end{nidan}\quad%
\begin{nidan}{#2}\end{nidan}}

\newcounter{toi} \newcounter{subtoi}[toi]
\newcounter{subsubtoi}[subtoi]

\everymath{\displaystyle}
\parindent=0zw \columnsep=3zw %\columnseprule=.4pt

\def\toi{\refstepcounter{toi}\par          % 問
  \makebox[2zw][r]{\Large\ajKaku{\value{toi}}\hspace{0.5zw}}%
  \hangindent=2zw}

\newcounter{subtoihyoji}
\def\subtoi{\refstepcounter{subtoi}\par    % 小問
  \setcounter{subtoihyoji}{\value{subtoi}}
	\ifnum\thesubtoi>20
		\addtocounter{subtoihyoji}{9873}
	\else
		\addtocounter{subtoihyoji}{8070}
	\fi
  \makebox[3.4zw][r]{\CID{\thesubtoihyoji}\hspace{1zw}}%
  \hangindent=3.4zw}

\newcounter{subsubtoihyoji}
\def\subsubtoi{\refstepcounter{subsubtoi}\par    % 小小問
  \setcounter{subsubtoihyoji}{\value{subsubtoi}}
  \addtocounter{subsubtoihyoji}{7554}
  \makebox[4.2zw][r]{\CID{\thesubsubtoihyoji}\hspace{1zw}}%
  \hangindent=4.2zw}

\font\kyocho=cmssbx10
\font\sskyocho=cmssbx10 at 8pt
\def\kotae#1{\hfill \textcolor{red}{\footnotesize\bf\(\boldsymbol{#1} \)}}

\begin{document}
%
%  ===========================以下本文===============================
%
\def\namae{%
	{\Large\bf 数学A \quad 授業プリント\# 16}\hfill 年\hskip8ex 組\hskip8ex 号 \hskip5ex
	
	\hfill 
	\underline{氏名\rule[-2ex]{0cm}{6ex}\hskip30ex}
}

\namae % 名前の出力

{\large\bf 自然数の和・平方数の和}

\bigskip


\begin{center}
\begin{minipage}[t]{0.4\textwidth}
\begin{shadebox}
\vspace*{2ex}

\mbox{}\hskip9zw\( 1+2+3+\cdots +n={\,1\, \over 2}n(n+1) \)\\
%\vspace*{1zh}
\end{shadebox}
\end{minipage}
\end{center}

\begin{center}
\begin{minipage}[t]{0.4\textwidth}
\begin{shadebox}
\vspace*{2ex}

\mbox{}\hskip7zw\( 1^2+2^2+3^3+\cdots +n^2={\,1\, \over 6}n(n+1)(2n+1) \)\\
%\vspace*{1zh}
\end{shadebox}
\end{minipage}
\end{center}


\fbox{例1}\hskip1zw \( 1+2+3+4+\cdots +20 \)を求めよ。

\fbox{解答}
\mbox{}\hskip0.75zw \( 1+2+3+4+\cdots +20=
{\,1\, \over 2}\times 20 \times (20+1) \)

\mbox{}\hskip14.35zw \( ={\,1\, \over 2}\times 20 \times 21 \)

\mbox{}\hskip14.35zw \( ={\,1\, \over 2}\times 20 \times 21 \)

\vskip1ex

\mbox{}\hskip14.35zw \( = \hskip8.5mm 10 \times 21 \)

\vskip1ex

\mbox{}\hskip14.35zw \( =210 \)


\vskip5zh



\fbox{例2}\hskip1zw \( 1^2+2^2+3^2+4^2+\cdots +15^2 \)を求めよ。

\fbox{解答}
\mbox{}\hskip0.75zw \( 1^2+2^2+3^2+4^2+\cdots +15^2=
{\,1\, \over 6}\times 15 \times (15+1) \times (2\times 15+1) \)

\mbox{}\hskip16.65zw \( ={\,1\, \over 6}\times 15 \times 16 \times 31 \)

\mbox{}\hskip16.65zw \( ={\,1\, \over 6}\times 15 \times 16 \times 31 \)

\vskip3ex

\mbox{}\hskip16.65zw \( =5\times 8 \times 31 \)

\vskip1ex

\mbox{}\hskip16.65zw \( = 1240 \)


\newpage


\toi 次の数列の和を求めよ。

\nidangumi{\subtoi \( 1+2+3+4+\cdots +15 \) }
			{\subtoi  \(  1+2+3+4+\cdots +22  \) }\vfill

\nidangumi{\subtoi \(1+2+3+4+\cdots +36\) }
			{\subtoi  \( 1+2+3+4+\cdots +55 \) }\vfill

\nidangumi{\subtoi \(1^2+2^2+3^2+4^2+\cdots +8^2\) }
			{\subtoi  \(1^2+2^2+3^2+4^2+\cdots +13^2\) }\vfill

\nidangumi{\subtoi \( 1^2+2^2+3^2+4^2+\cdots +21^2 \) }
			{\subtoi  \(1^2+2^2+3^2+4^2+\cdots +45^2\) }\vfill

\nidangumi{\subtoi \( 11+12+13+14+\cdots +36\) }
			{\subtoi  \( 16^2+17^2+18^2+19^2+20^2\) }\vfill



\end{document}
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