# Difference between revisions of "User:Pifinity"

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= pifinity's problem set = | = pifinity's problem set = | ||

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<b>Welcome to pifinity's problem set! This page contains problems created by pifinity. Please do not edit this page.</b> | <b>Welcome to pifinity's problem set! This page contains problems created by pifinity. Please do not edit this page.</b> | ||

− | + | == Finalized Problems == | |

− | == Problem | + | ==== Problem I ==== |

− | + | Evaluate <math>\frac{48}{32} + \frac{24}{16}.</math> | |

− | <cite><sub>This page has | + | ==== Solution I ==== |

+ | Simplify; no denominator conversion needed: <cmath>\frac{48}{32} + \frac{24}{16}=\frac{3}{2} + \frac{3}{2} = \boxed{3}.</cmath> | ||

+ | |||

+ | |||

+ | == Unconfirmed Problems == | ||

+ | ==== Problem ==== | ||

+ | What is the <math>``\,?\,"</math> term? <cmath>(100-96)+100+(96-92)+94+(92-88)+88+\hdots+?+(4-0)</cmath> | ||

+ | ==== Solution ==== | ||

+ | We first wish to find the number of "telescoping" terms. The first one is <math>100</math> minus something, so we can label it <math>\tfrac{100}4=25.</math> The last term is <math>4</math> minus something, so we can label it <math>\tfrac{4}4=1.</math> Thus there are <math>25-(4-1)=22</math> terms. Therefore, there are 21 gaps between the terms in parentheses. Since there is one less of the terms not in parentheses, then there are 20 gaps between these terms in total. And since <math>6</math> is subtracted each time for the terms not in parentheses, then we have to subtract <math>6\times20=120</math> from <math>100</math>: <cmath>100-120=\boxed{-20}.</cmath> | ||

+ | == Unsolved Problems == | ||

+ | <b>No Unsolved Problems.</b> | ||

+ | |||

+ | = = | ||

+ | <cite><sub>This page has 12 official edits. The last edit was on 12:09, May 18, 2018.</sub></cite> |

## Latest revision as of 14:09, 18 May 2018

## Contents

# pifinity's problem set

**Welcome to pifinity's problem set! This page contains problems created by pifinity. Please do not edit this page.**

## Finalized Problems

#### Problem I

Evaluate

#### Solution I

Simplify; no denominator conversion needed:

## Unconfirmed Problems

#### Problem

What is the term?

#### Solution

We first wish to find the number of "telescoping" terms. The first one is minus something, so we can label it The last term is minus something, so we can label it Thus there are terms. Therefore, there are 21 gaps between the terms in parentheses. Since there is one less of the terms not in parentheses, then there are 20 gaps between these terms in total. And since is subtracted each time for the terms not in parentheses, then we have to subtract from :

## Unsolved Problems

**No Unsolved Problems.**

_{This page has 12 official edits. The last edit was on 12:09, May 18, 2018.}