Problem 66 - Beginning and Intermediate Algebra

Question

Find the indicated term of each binomial expansion. Show that \(\left(\begin{array}{l}n \\ 1\end{array}\right)=n\) for any positive integer \(n\)

Step-by-Step Solution

Answer

We showed that \(\left(\begin{array}{l}n \\\ 1\end{array}\right)=n\) for any positive integer \(n\) using the n choose r formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). By substituting r with 1 and simplifying the expression, we found that \[\binom{n}{1} = \frac{n!}{1!(n-1)!} = \frac{n!}{(n-1)!} = n\].

Step 1: Write the formula for n choose r

The formula for n choose r is: \[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]

Step 2: Substitute r with 1

We are given that we need to find \(\left(\begin{array}{l}n \\\ 1\end{array}\right)\), so we'll substitute r with 1: \[\binom{n}{1} = \frac{n!}{1!(n-1)!}\]

Step 3: Simplify the expression

Now, we'll simplify the expression, knowing that \(0! = 1\) and \(1! = 1\), we get: \[\binom{n}{1} = \frac{n!}{1!(n-1)!} = \frac{n!}{(n-1)!}\]

Step 4: Cancel out common terms

We can write the n! as a product of all positive integers up to n: \(\underbrace{n \cdot \ldots \cdot 3 \cdot 2 \cdot 1}_{n!}\) Likewise, the (n-1)! can be written as the product of all positive integers up to (n-1): \(\underbrace{(n-1) \cdot \ldots \cdot 2 \cdot 1}_{(n-1)!}\) Now the expression becomes: \[\binom{n}{1} = \frac{\underbrace{n \cdot \ldots \cdot 3 \cdot 2 \cdot 1}_{n!}}{\underbrace{(n-1) \cdot \ldots \cdot 2 \cdot 1}_{(n-1)!}}\] We can cancel out the common terms: \[\binom{n}{1} = n\]

Step 5: Conclusion

Therefore, we have shown that \(\left(\begin{array}{l}n \\\ 1\end{array}\right)=n\) for any positive integer \(n\).