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Here is my part of script.

\Rightarrow (n+1)^2+5(n+1)+1\nonumber
&=& n^2+2n+1+5n+5+1\nonumber\\
&=& (n^2+5n+1)+2n+6\nonumber\\
&=& 2k+2n+6 \cdots\text{for some $k\in\N$}\nonumber\\
&=& 2(k+n+3)\nonumber

Here, I used \nonumber but I'm being tired to put them everytime I type this thing...

Is there a way to turn off the autonumbering in equnarray?

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@Mico How did you make the "script" part so nice? – User Aug 28 '14 at 20:53
I highlighted the lines in question and clicked on the {} "button" in the row above the editing window. Doing so indents the highlighted lines by four spaces, pretty-printing them in the process. – Mico Aug 28 '14 at 20:55
up vote 13 down vote accepted

Use eqnarray* instead of eqnarray to get an unnumbered multiline equation structure.

However, you really shouldn't be using either eqnarray or eqnarray* -- both environments have been seriously deprecated for many years. Instead, use align and align*, respectively. For more in-depth references on this subject, see the posting eqnarray vs align.

One of the most annoying shortcomings of eqnarray or eqnarray* is the poor spacing around the alignment point (usually =): They insert far more much whitespace on either side of the alignment point than any of the other TeX and LaTeX methods do for typesetting equations. In the code below, compare the excessive amount of whitespace around the = symbols in the first structure, generated with eqnarray*, with the normal amount in the second structure, which is generated with align*.

If nothing else can convince you to give up on eqnarray, consider the following: If you switch from eqnarray to align, you'll save yourself some typing -- the word align is shorter than eqnarray -- and you need to type only one & symbol to indicate the alignment point. :-)

enter image description here


With \verb+eqnarray*+:
\Rightarrow (n+1)^2+5(n+1)+1
&=& n^2+2n+1+5n+5+1\\
&=& (n^2+5n+1)+2n+6\\
&=& 2k+2n+6 \qquad \text{for some $k\in\N$}\\
&=& 2(k+n+3)

With \verb+align*+:
\Rightarrow (n+1)^2+5(n+1)+1
&= n^2+2n+1+5n+5+1\\
&= (n^2+5n+1)+2n+6\\
&= 2k+2n+6 \qquad \text{for some $k\in\N$}\\
&= 2(k+n+3)
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