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<HTML><HEAD><TITLE>About Arithmetic Series Summation</TITLE></HEAD><BODY bgcolor="#FFFFFF"> <a href="../index.html" target="_top"><img src="../icons/usins.gif" alt="Logo" align=right></a><b>Data Structures and Algorithms with Object-Oriented Design Patterns in Python</b><br><A NAME="tex2html1794" HREF="page52.html"><IMG WIDTH=37 HEIGHT=24 ALIGN=BOTTOM ALT="next" SRC="../icons/next_motif.gif"></A> <A NAME="tex2html1792" HREF="page49.html"><IMG WIDTH=26 HEIGHT=24 ALIGN=BOTTOM ALT="up" SRC="../icons/up_motif.gif"></A> <A NAME="tex2html1786" HREF="page50.html"><IMG WIDTH=63 HEIGHT=24 ALIGN=BOTTOM ALT="previous" SRC="../icons/previous_motif.gif"></A> <A NAME="tex2html1796" HREF="page611.html"><IMG WIDTH=43 HEIGHT=24 ALIGN=BOTTOM ALT="index" SRC="../icons/index_motif.gif"></A> <BR><HR><H2><A NAME="SECTION002220000000000000000">About Arithmetic Series Summation</A></H2><P>The series, <IMG WIDTH=77 HEIGHT=22 ALIGN=MIDDLE ALT="tex2html_wrap_inline58119" SRC="img141.gif" >,is an <em>arithmetic series</em><A NAME=914> </A>and the summation<P> <IMG WIDTH=284 HEIGHT=44 ALIGN=BOTTOM ALT="displaymath58117" SRC="img142.gif" ><P>is called the<em>arithmetic series summation</em><A NAME=918> </A>.<P>The summation can be solved as follows:First, we make the simple variable substitution <I>i</I>=<I>n</I>-<I>j</I>: <P><A NAME="eqnmodelarith"> </A> <IMG WIDTH=500 HEIGHT=220 ALIGN=BOTTOM ALT="eqnarray919" SRC="img143.gif" ><P>Note that the term in the first summation in Equation <A HREF="page51.html#eqnmodelarith"><IMG ALIGN=BOTTOM ALT="gif" SRC="../icons/cross_ref_motif.gif"></A>is independent of <I>j</I>.Also, the second summation is identical to the left hand side.Rearranging Equation <A HREF="page51.html#eqnmodelarith"><IMG ALIGN=BOTTOM ALT="gif" SRC="../icons/cross_ref_motif.gif"></A>, and simplifying gives <P> <IMG WIDTH=500 HEIGHT=150 ALIGN=BOTTOM ALT="eqnarray938" SRC="img144.gif" ><P><P>There is, of course, a simpler way to arrive this answer.Consider the series, <IMG WIDTH=97 HEIGHT=22 ALIGN=MIDDLE ALT="tex2html_wrap_inline58125" SRC="img145.gif" >,and suppose <I>n</I> is even.The sum of the first and last element is <I>n</I>+1.So too is the sum of the second and second-last element,and the third and third-last element, etc.,and there are <I>n</I>/2 such pairs.Therefore, <IMG WIDTH=98 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline58133" SRC="img146.gif" >.<P>And if <I>n</I> is odd, then <IMG WIDTH=102 HEIGHT=21 ALIGN=MIDDLE ALT="tex2html_wrap_inline58137" SRC="img147.gif" >,where <I>n</I>-1 is even.So we can use the previous result for <IMG WIDTH=32 HEIGHT=21 ALIGN=MIDDLE ALT="tex2html_wrap_inline58141" SRC="img148.gif" > to get <IMG WIDTH=201 HEIGHT=29 ALIGN=MIDDLE ALT="tex2html_wrap_inline58143" SRC="img149.gif" >.<P><HR><A NAME="tex2html1794" HREF="page52.html"><IMG WIDTH=37 HEIGHT=24 ALIGN=BOTTOM ALT="next" SRC="../icons/next_motif.gif"></A> <A NAME="tex2html1792" HREF="page49.html"><IMG WIDTH=26 HEIGHT=24 ALIGN=BOTTOM ALT="up" SRC="../icons/up_motif.gif"></A> <A NAME="tex2html1786" HREF="page50.html"><IMG WIDTH=63 HEIGHT=24 ALIGN=BOTTOM ALT="previous" SRC="../icons/previous_motif.gif"></A> <A NAME="tex2html1796" HREF="page611.html"><IMG WIDTH=43 HEIGHT=24 ALIGN=BOTTOM ALT="index" SRC="../icons/index_motif.gif"></A> <P><ADDRESS><img src="../icons/bruno.gif" alt="Bruno" align=right><a href="../copyright.html">Copyright © 2003</a> by <a href="../signature.html">Bruno R. Preiss, P.Eng.</a> All rights reserved.</ADDRESS></BODY></HTML>⌨️ 快捷键说明
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