Multi Numerica (Jan 2016)
Author:
vivek378
Last Updated:
9 years ago
License:
Creative Commons CC BY 4.0
Abstract:
Bad Math Journal
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
Bad Math Journal
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\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}