Summation Subtraction Formatting

Question

I am trying to stack two summations -- one on top of the other -- to illustarte term cancellation that takes place when subtracting the two sums. I want the summations to align at the first "=" sign, be centered on the page, and also be underlined. Here is as close as I've gotten:

$$\sum_{i=1}^{n}a_i =
        a_1+a_2+\ldots+a_{n-2}+a_{n-1}+a_n =
        \left[\frac{n(n+1)}{2}\right]^4$$

$$-\underline{\sum_{i=1}^{n-1}a_i = 
        a_1+a_2+\ldots+a_{n-2}+a_{n-1} =
        \left[\frac{n(n-1)}{2}\right]^4}$$

$$a_n = \left[\frac{n(n+1)}{2}\right]^4 -
          \left[\frac{n(n-1)}{2}\right]^4$$

Any ideas???

Answer 1

Your question is most likely a duplicate of this question.

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs}
\begin{document}
% from https://tex.stackexchange.com/questions/23014/horizontal-line-separating-steps-in-aligned-equation
\begin{equation*}
\begin{aligned}
\sum_{i=1}^{n}a_i &=
        a_1+a_2+\ldots+a_{n-2}+a_{n-1}+a_n =
        \left[\frac{n(n+1)}{2}\right]^4\\
-\sum_{i=1}^{n-1}a_i &= 
        -\left(a_1+a_2+\ldots+a_{n-2}+a_{n-1}\right) =
        -\left[\frac{n(n-1)}{2}\right]^4\\
\midrule
a_n &= \left[\frac{n(n+1)}{2}\right]^4 -
          \left[\frac{n(n-1)}{2}\right]^4
\end{aligned}         
\end{equation*}
\end{document}

Answer 2

One possibility is to use align* with the amsmath package. This keeps the spacing correct throughout, unlike the spacing created by making an array. You also have the option to number the rows by removing the *. Alignment is maintained using a \phantom command.

The horizontal line here is inserted as a \rule using \intertext. This can be automated using the booktabs package as others have mentioned.

Code is here:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{align*}
\sum_{i=1}^{n}a_i = a_1+a_2+\cdots+a_{n-2}+a_{n-1}+a_n &= \left[\frac{n(n+1)}{2}\right]^4\\
-\sum_{i=1}^{n-1}a_i = a_1+a_2+\cdots+a_{n-2}+a_{n-1}\phantom{{}+a_n} &= \left[\frac{n(n-1)}{2}\right]^4\\
\intertext{\centering\rule{4.6in}{1pt}}
 a_n &=\left[\frac{n(n+1)}{2}\right]^4-\left[\frac{n(n-1)}{2}\right]^4
\end{align*}

\end{document}

I suppose \cdots vs. \ldots is a personal preference.

Answer 3

Another variation on the theme; I find that either you add a minus signs in front of each term in the second equation or in front of none. Just one minus sign is incorrect in all respects.

You can instead use words in order to help the reader in understanding what’s going on.

\documentclass{article}
\usepackage{amsmath,booktabs}

\begin{document}

In order to compute $a_n$, subtract the
second equation from the first one:
\begin{equation*}
\begin{gathered}
\begin{alignedat}{2}
\sum_{i=1}^{n}a_i
 & = a_1+a_2+\dots+a_{n-2}+a_{n-1}+a_n
&& = \left[\frac{n(n+1)}{2}\right]^4
\\
\sum_{i=1}^{n-1}a_i
 & = a_1+a_2+\dots+a_{n-2}+a_{n-1}
&& = \left[\frac{n(n-1)}{2}\right]^4
\end{alignedat}
\\
\midrule
a_n = \left[\frac{n(n+1)}{2}\right]^4
    - \left[\frac{n(n-1)}{2}\right]^4
\end{gathered}
\end{equation*}

\end{document}

Note. \dots should be used between alike symbols, in this case +; amsmath will take care of what kind of dots is needed. In case the dots are trailing, use \dotsb (like between binary symbols), \dotsm (like for multiplication), \dotsi (like between integrals), \dotsc (like between punctuation).

A different way to present the same argument.

\documentclass{article}
\usepackage{amsmath,booktabs}

\begin{document}

We have proved that, for every $n$,
\begin{equation*}
\sum_{i=1}^{n}a_i = \left[\frac{n(n+1)}{2}\right]^4.
\end{equation*}
Thus, for $n\ge1$, we can say
\begin{equation*}
\sum_{i=1}^{n-1}a_i = \left[\frac{n(n-1)}{2}\right]^4
\end{equation*}
which has the consequence that
\begin{equation*}
\left[\frac{n(n+1)}{2}\right]^4=
\sum_{i=1}^{n}a_i = 
a_n+\sum_{i=1}^{n-1}a_i = 
a_n+ \left[\frac{n(n-1)}{2}\right]^4.
\end{equation*}
Hence, by comparing the first and the last terms,
\begin{equation*}
a_n=\left[\frac{n(n+1)}{2}\right]^4-\left[\frac{n(n-1)}{2}\right]^4.
\end{equation*}

\end{document}

Answer 4

All answers here are LaTeX oriented but I am not able to read, that LaTeX is required from your answer. So, here is plain TeX solution:

In order to compute $a_n$, subtract the
second equation from the first one:
$$\eqalign{
   \sum_{i=1}^n a_i & = 
     a_1+a_2+\cdots+a_{n-2}+a_{n-1}+a_n = \left[n(n+1)\over 2\right]^4
   \cr
   \sum_{i=1}^{n-1}a_i & = 
     a_1+a_2+\cdots+a_{n-2}+a_{n-1}\phantom{{}+a_n}= \left[n(n-1)\over 2\right]^4
   \cr \noalign{\medskip\hrule\medskip}
   a_n & = \left[n(n+1)\over 2\right]^4 - \left[n(n-1)\over 2\right]^4
}
$$

\bye

Answer 5

I think the idea of this is to align term to term both sums, so one can show subtraction of summations more properly. For this, one can use an array.

By the way, to pursuing math accuracy, I included the needed - sign before all members of the equations in the second line (not only in the first member) and wrapped into parentheses the expanded sum.

\documentclass{article}
\usepackage{amsmath}
  \renewcommand{\arraycolsep}{1pt}


\begin{document}

\subsection*{In-line math display}

\[
\begin{array}{rcrl}
\sum_{i=1}^{n}a_i       &  =  &   a_1 + a_2 + \cdots + a_{n-2} + a_{n-1} + a_n          & = \left[\frac{n(n+1)}{2}\right]^4 \\[2ex]
-\sum_{i=1}^{n-1}a_i    &  =  & - (a_1 + a_2 + \cdots + a_{n-2} + a_{n-1})\hphantom{mm} & = -\left[\frac{n(n-1)}{2}\right]^4 \\[2ex]\hline\\[-2ex]
                        &     &                                                 a_n     & = \left[\frac{n(n+1)}{2}\right]^4-\left[\frac{n(n-1)}{2}\right]^4
\end{array}
\]

\bigskip

\subsection*{Displaymath}
\[
\begin{array}{rcrl}
\displaystyle\sum_{i=1}^{n}a_i    &  =  &   a_1 + a_2 + \cdots + a_{n-2} + a_{n-1} + a_n          & = \left[\dfrac{n(n+1)}{2}\right]^4 \\[4ex]
\displaystyle-\sum_{i=1}^{n-1}a_i &  =  & -(a_1 + a_2 + \cdots + a_{n-2} + a_{n-1})\hphantom{mm}  & = -\left[\dfrac{n(n-1)}{2}\right]^4 \\[4ex]\hline\\[-1ex]
                                  &     &                                            a_n          & = \left[\dfrac{n(n+1)}{2}\right]^4-\left[\dfrac{n(n-1)}{2}\right]^4
\end{array}
\]

\end{document}