Multi Numerica (Jan 2016)
Forfatter:
vivek378
Sidst opdateret:
9 år siden
Licens:
Creative Commons CC BY 4.0
Resumé:
Bad Math Journal
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
Bad Math Journal
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
\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}