Multi Numerica (Jan 2016)
Author:
vivek378
Last Updated:
9 years ago
License:
Creative Commons CC BY 4.0
Abstract:
Bad Math Journal
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\documentclass[a4paper]{article}
\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage[colorinlistoftodos]{todonotes}
\title{Multi Numerica (Jan 2016)}
\author{Vivek Alumootil}
\begin{document}
\maketitle
\vspace{15mm}
\section{Geometric Power Series Function}
Let $f(n,m,z,a) = n^a + n^{a+z} +... n^{a+(m-1)z} = \sum\limits_{i=0}^m n^{a+(i-1)z} = s$\\\\
$=s = \sum\limits_{i=0}^m n^{a+(i-1)z}$ \\\\
$=sn^z=\sum\limits_{i=0}^m n^{a+iz}$ \\\\
$=sn^z-s=\sum\limits_{i=0}^m n^{a+iz}-\sum\limits_{i=0}^m n^{a+(i-1)z}$ \\\\
$=sn^z-s = n^{a+mz}-n^a$\\\\
$=s(n^z-1) = n^{mz+a}-n^{a})$\\\\
$=s=\frac{n^{mz+a}-n^{a}}{n^z-1}$\\\\
$\boxed{f(n,m,z,a)=\frac{n^{mz+a}-n^{a}}{n^z-1}}$ \\\\\\\\
\section{The Sum of the First n Hexagonal Numbers}
$\sum\limits_{i=1}^n \frac{(2i-1)(2i)}{2}$\\\\
$=\sum\limits_{i=1}^n i(2i-1)$\\\\
$=\sum\limits_{i=1}^n 2i^2-i$\\\\
$=\sum\limits_{i=1}^n 2i^2 - \sum\limits_{i=1}^n i$\\\\
$=\frac{n(n+1)(2n+1)}{3}-\frac{n(n+1)}{2}$ \\\\
$= \frac{2n(n+1)(2n+1)-3n(n+1)}{6}$ \\\\
$=\frac{n(n+1)(4n+2-3)}{6}$ \\\\
$=\frac{n(n+1)(4n-1)}{6}$ \\\\
$\boxed{\frac{n(n+1)(4n-1)}{6}}$ \\\\\\\\
\section{The Sum of the First n s-gonal Numbers}
Note that xth s-gonal number P(s,x) is \\
$P(s,x) = \frac{x^2(s-2)-x(s-4)}{2}$\\\\\\
Let $f(n,s)= \sum\limits_{i=1}^n \frac{i^2(s-2)-i(s-4)}{2}$ \\\\
$=f(n,s)=\sum\limits_{i=1}^n \frac{i^2(s-2)}{2}-\frac{i(s-4)}{2}$ \\\\
$=f(n,s)=\frac{n(n+1)(2n+1)}{6} \frac{s-2}{2} - \frac{n(n+1)}{2} \frac{s-4}{2}$ \\\\
$=f(n,s) = \frac{n(n+1)(2n+1)(s-2)}{12}-\frac{n(n+1)(s-4)}{4}$ \\\\
$=f(n,s) = \frac{n(n+1)(2n+1)(s-2)-3n(n+1)(s-4)}{12}$ \\\\
$=f(n,s) = \frac{n(n+1)((2n+1)(s-2)-3(s-4))}{12}$ \\\\
$\boxed{f(n,s) = \frac{n(n+1)((2n+1)(s-2)-3(s-4))}{12}}$ \\\\\\\\
\section{The n Products of the Sum of Squares}
$\prod\limits_{j=1}^n (\sum\limits_{i=1}^j i^2)$ \\\\
$=\prod\limits_{j=1}^n \frac{j(j+1)(2j+1)}{6}$ \\\\
$=\frac{\prod\limits_{j=1}^n j(j+1)(2j+1)}{6^n}$ \\\\
$=\frac{n!(n+1)!\prod\limits_{j=1}^n (2j+1)}{6^n}$ \\\\
$=\frac{n!(n+1)!\frac{(2n+1)!}{2}}{2^{n-1}n!6^n}$ \\\\
$=\frac{n!(n+1)!(2n+1)!}{2^nn!6^n}$\\\\
$=\frac{(n+1)!(2n+1)!}{12^n}$ \\\\
$\boxed{\frac{(n+1)!(2n+1)!}{12^n}}$ \\\\\\\\
\end{document}