## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.1 - Sequences and Summation Notation - Concept and Vocabulary Check - Page 1048: 7

#### Answer

$\left( n+3 \right)$.

#### Work Step by Step

We know that $n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\cdots 3\times 2\times 1$. Now to evaluate fractions with factorials in the numerator and the denominator, we have to reduce the fraction before performing the multiplications. Here, for $\frac{\left( n+3 \right)!}{\left( n+2 \right)!}$ write $\left( n+3 \right)!$as$\left( n+3 \right)\left( n+2 \right)!$. After that, we can divide both the numerator and the denominator by the common factor and therefore we can reduce the multiplication. We are given that, $\frac{\left( n+3 \right)!}{\left( n+2 \right)!}$ Now we can write it as $\frac{(n+3)(n+2)!}{(n+2)!}$. Now, divide the numerator and denominator by $(n+2)!$ to get $\frac{\left( n+3 \right)!}{\left( n+2 \right)!}=\left( n+3 \right)$.

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