Problem 65 Show that $$ \sum_{k=0}^{n} ... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

2 Edition

Chapter 7: Problem 65

Show that $$ \sum_{k=0}^{n} \frac{n !}{k !(n-k) !}=2^{n} $$ for every positive integer $n$. [Hint: Expand $(1+1)^{n}$ using the Binomial Theorem.]

Short Answer

Expert verified

Recall the Binomial Theorem: $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}$. Apply the theorem to expand $(1+1)^n$, which simplifies to $\sum_{k=0}^{n} \binom{n}{k}$. We can compare the given expression, $\sum_{k=0}^{n} \frac{n !}{k !(n-k) !}$, to the expansion of $(1+1)^n$ and notice that the expression inside the summation is the same as the binomial coefficient. Thus, the given expression is the same as the expansion of $(1+1)^n$. Since $(1+1)^n = 2^n$, we can conclude that $\sum_{k=0}^{n} \frac{n !}{k !(n-k) !} = 2^n$ for every positive integer n.

Step by step solution

Recall the Binomial Theorem

The Binomial Theorem states that for any positive integer n and real numbers a and b: \[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\] where $\binom{n}{k}$ denotes the binomial coefficient, which is given by: \[\binom{n}{k} = \frac{n!}{k!(n-k) !}\]

Apply the Binomial Theorem to expand $(1+1)^n$

We can apply the Binomial Theorem to expand $(1+1)^n$: \[(1+1)^n = \sum_{k=0}^{n} \binom{n}{k} 1^{n-k} 1^{k}\] Since 1 raised to any power is still 1, the expression simplifies to: \[(1+1)^n = \sum_{k=0}^{n} \binom{n}{k}\]

Compare the given expression

Now, let's compare the given expression to the expansion of $(1+1)^n$: \[\sum_{k=0}^{n} \frac{n !}{k !(n-k) !} = \sum_{k=0}^{n} \binom{n}{k}\] Notice that the expression inside the summation is the same as the binomial coefficient. Therefore, the given expression is the same as the expansion of $(1+1)^n$.

Simplify the expansion of $(1+1)^n$

Simplify the left-hand side of the equation: \[(1+1)^n = 2^n\]

Conclude the proof

Since we have shown that the given expression is the expansion of $(1+1)^n$ and that that expansion is equal to $2^n$, we can conclude that: \[\sum_{k=0}^{n} \frac{n !}{k !(n-k) !} = 2^n\] for every positive integer n.

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91影视

Show that $$ \sum_{k=0}^{n} \frac{n !}{k !(n-k) !}=2^{n} $$ for every positive integer \(n\). [Hint: Expand \((1+1)^{n}\) using the Binomial Theorem.]

Short Answer

Step by step solution

Recall the Binomial Theorem

Apply the Binomial Theorem to expand \((1+1)^n\)

Compare the given expression

Simplify the expansion of \((1+1)^n\)

Conclude the proof

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