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I am using landscape page in LaTeX. I am writing the below but instead of getting in the first column, and they should, they go to the 2nd column?

\documentclass[landscape,twocolumn,letterpaper]{report}
\usepackage[landscape,twocolumn]{geometry}
%\usepackage{tikz-cd}
%\usepackage{fullpage}
\usepackage{amssymb}
%\usepackage{amused}
\usepackage{mathrsfs}
%\usepackage{eureka}
%\def\principaladviser#1{\gdef\@principaladviser{#1}}
\usepackage[centertags]{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
%\usepackage[all]{xy}
\usepackage{epsfig}
\usepackage{graphicx}
\usepackage{amsthm}
%\usepackage{breqn}
%\usepackage{verbatim}
%\usepackage{apst-all}
%\usepackage{xy-pic}
\usepackage{amssymb,latexsym}
\usepackage{amsfonts,amsmath}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{textcomp}
%\usepackage{gensymb}
\usepackage{amsmath,amssymb}
%\usepackage{dsfont}\let\mathbb\mathds
\usepackage{latexsym}
\usepackage{amssymb}
\usepackage{amsfonts}
%\usepackage{fancyhdr}
\usepackage[arrow,frame,matrix]{xy}
\usepackage{amsmath}
%\usepackage{lipsum}
%\usepackage{eqlist}
\usepackage{fixltx2e}
%\usepackage[demo]{graphicx}
%\usepackage{mwe}
%\usepackage{breqn}
\renewcommand{\baselinestretch}{1.5}
\begin{document}
$$4\sum_{j=1}^{j=n}j^3+6\sum_{j=1}^{j=n}j^2+4\sum_{j=1}^{j=n}j\\+\sum_{j=1}^{j=n}1=b_{n+1}-b_{1}$$
\begin{align*}\Rightarrow &4\sum_{j=1}^{j=n}j^3+6\sum_{j=1}^{j=n}j^2+4\sum_{j=1}^{j=n}j+\sum_{j=1}^{j=n}1=(n+1)^4-1\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3+6\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}+n=(n+1)^4-1\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)^4-1-n-2n(n+1)-n(n+1)(2n+1)\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)^4-(n+1)-2n(n+1)-n(n+1)(2n+1)\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)[(n+1)^3-n(2n+1)-2n-1]\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)[n^3+3n^2+3n+1-2n^2-n-2n-1]\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)(n^3+n^2)\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=n^2(n+1)(n+1)\\
\Rightarrow &\sum_{j=1}^{j=n}j^3=\frac{n^2(n+1)(n+1)}{4}\\
\Rightarrow &\sum_{j=1}^{j=n}j^3=\big(\frac{n(n+1)}{2}\big)^2\\
\Rightarrow &1^3+2^2+3^3+...+n^3=\bigg[\frac{n(n+1)}{2}\bigg]^2\\
\end{align*}

\end{document} 
  • @moewe Brother I tried so much to write the code in edited form, can you tell me how to write code in this correct form when posting question! – Noor Aslam Mar 26 at 6:22
  • 1
    Select everything you want to mark up as code and press the {...} button in the editor. That will indent the selected material by four spaces so it is recognised as code (of course you can also add the spaces manually). Alternatively, you could add three backticks (```) just above and below the codeblock. See meta.stackexchange.com/q/22186 for a more thorough explanation with pictures. – moewe Mar 26 at 6:31
  • 1
    Aparently, contents of the align* environment don't fit into the left column so they are moved to the right column. If you want the align* environment to start in the left column and continue in the right column, you can use \allowdisplaybreaks. – leandriis Mar 26 at 8:31
  • BTW, some of your fonts are obsolete and many are loaded multiple times. – John Kormylo Mar 26 at 14:04
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I would suggest to use eqnarray* environment. Also I've defined two columns within document not for whole document at the beginning. And aligned under eqnarray* environment. So please note I've added \usepackage{eqnarray}.

Take a look at my example:

\documentclass[landscape,letterpaper]{report}
\usepackage[landscape,showframe]{geometry}
\usepackage{amsmath}
\usepackage{eqnarray}
\usepackage{multicol}
\renewcommand{\baselinestretch}{1.5}
\begin{document}
\begin{multicols}{2}
\begin{eqnarray*}
\begin{aligned}
&4\sum_{j=1}^{j=n}j^3+6\sum_{j=1}^{j=n}j^2+4\sum_{j=1}^{j=n}j+\sum_{j=1}^{j=n}1=b_{n+1}-b_{1}...\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3+6\sum_{j=1}^{j=n}j^2+4\sum_{j=1}^{j=n}j+\sum_{j=1}^{j=n}1=(n+1)^4-1\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3+6\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}+n=(n+1)^4-1\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)^4-1-n-2n(n+1)-n(n+1)(2n+1)\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)^4-(n+1)-2n(n+1)-n(n+1)(2n+1)\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)[(n+1)^3-n(2n+1)-2n-1]\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)[n^3+3n^2+3n+1-2n^2-n-2n-1]\\
\Rightarrow &4\sum_{j=1}^{j=n}j^3=(n+1)(n^3+n^2)\\
\Rightarrow &\sum_{j=1}^{j=n}j^3=n^2(n+1)(n+1)\\
\Rightarrow &\sum_{j=1}^{j=n}j^3=\frac{n^2(n+1)(n+1)}{4}\\
\Rightarrow &\sum_{j=1}^{j=n}j^3=\big(\frac{n(n+1)}{2}\big)^2\\
\Rightarrow &1^3+2^2+3^3+...+n^3=\bigg[\frac{n(n+1)}{2}\bigg]^2
\end{aligned}
\end{eqnarray*}
\end{multicols}
\end{document} 

Is it what you want to get?

enter image description here

If not you can split equation array and second part will appear on right side of the page.

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  • Where did you get the package eqarray from? – leandriis Mar 26 at 8:29
  • My bad. I made mistake being in hurry. eqnarray not eqarray – Moonray Mar 26 at 16:15
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Your text doesn't quite fit into a default page. You either need to make the text area larger or use a smaller font size.

BTW, I moved the first equation all the way left and the others (collectively) all the way right. Also, it is better to put & to the left of a binary operator than to the right, otherwise you should use \null&.

\documentclass[landscape,twocolumn,letterpaper]{report}
\usepackage[margin=1in,showframe]{geometry}
%\usepackage{tikz-cd}
%\usepackage{fullpage}
%\usepackage{amssymb}
%\usepackage{amused}
\usepackage{mathrsfs}
%\usepackage{eureka}
%\def\principaladviser#1{\gdef\@principaladviser{#1}}
\usepackage[centertags]{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
%\usepackage[all]{xy}
\usepackage{epsfig}
\usepackage{graphicx}
%\usepackage{amsthm}
%\usepackage{breqn}
%\usepackage{verbatim}
%\usepackage{apst-all}
%\usepackage{xy-pic}
%\usepackage{amssymb,latexsym}
%\usepackage{amsfonts,amsmath}
%\usepackage[utf8]{inputenc}
%\usepackage[T1]{fontenc}
%\usepackage{textcomp}
%\usepackage{gensymb}
%\usepackage{amsfonts}
%\usepackage{fancyhdr}
\usepackage[arrow,frame,matrix]{xy}
%\usepackage{amsmath}
%\usepackage{lipsum}
%\usepackage{eqlist}
\usepackage{fixltx2e}
%\usepackage[demo]{graphicx}
%\usepackage{mwe}
%\usepackage{breqn}
\renewcommand{\baselinestretch}{1.5}% Really?

\columnseprule=0.5pt% show column border

\begin{document}
\begin{flalign*}
\sum_{j=1}^{j=n}j^3
  +6\sum_{j=1}^{j=n}j^2
  +4\sum_{j=1}^{j=n}j 
  +\sum_{j=1}^{j=n}1=b_{n+1}-b_{1} &&&&
\end{flalign*}\vspace{\dimexpr -\abovedisplayskip-\belowdisplayskip}
\begin{flalign*}
&&&\Rightarrow 4\sum_{j=1}^{j=n}j^3+6\sum_{j=1}^{j=n}j^2+4\sum_{j=1}^{j=n}j+\sum_{j=1}^{j=n}1=(n+1)^4-1\\
&&&\Rightarrow 4\sum_{j=1}^{j=n}j^3+6\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}+n=(n+1)^4-1\\
&&&\Rightarrow 4\sum_{j=1}^{j=n}j^3=(n+1)^4-1-n-2n(n+1)-n(n+1)(2n+1)\\
&&&\Rightarrow 4\sum_{j=1}^{j=n}j^3=(n+1)^4-(n+1)-2n(n+1)-n(n+1)(2n+1)\\
&&&\Rightarrow 4\sum_{j=1}^{j=n}j^3=(n+1)[(n+1)^3-n(2n+1)-2n-1]\\
&&&\Rightarrow 4\sum_{j=1}^{j=n}j^3=(n+1)[n^3+3n^2+3n+1-2n^2-n-2n-1]\\
&&&\Rightarrow 4\sum_{j=1}^{j=n}j^3=(n+1)(n^3+n^2)\\
&&&\Rightarrow 4\sum_{j=1}^{j=n}j^3=n^2(n+1)(n+1)\\
&&&\Rightarrow \sum_{j=1}^{j=n}j^3=\frac{n^2(n+1)(n+1)}{4}\\
&&&\Rightarrow \sum_{j=1}^{j=n}j^3=\big(\frac{n(n+1)}{2}\big)^2\\
&&&\Rightarrow 1^3+2^2+3^3+...+n^3=\bigg[\frac{n(n+1)}{2}\bigg]^2
\end{flalign*}

\end{document} 
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