Multi Numerica (Jan 2016)
Author:
vivek378
Last Updated:
9 лет назад
License:
Creative Commons CC BY 4.0
Аннотация:
Bad Math Journal
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
Bad Math Journal
\begin
Discover why over 20 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}