Problem 31
Question
Using the binomial theorem, prove each. \(\sum_{r=0}^{n} 2^{r}\left(\begin{array}{l}n \\ r\end{array}\right)=3^{n}\)
Step-by-Step Solution
Verified Answer
By using the binomial theorem with appropriate values for a and b (a=1 and b=2), we can expand \((1+2)^n = 3^n\), which simplifies to \(\sum_{r=0}^{n} \binom{n}{r} 2^r = 3^n\). This matches the expression we needed to prove, confirming that \(\sum_{r=0}^{n} 2^{r}\binom{n}{r} = 3^{n}\).
1Step 1: Identify the appropriate values for a and b
As we want to prove the given equation using the binomial theorem, let's go for a = 1 and b = 2, so that it has the power of 2 which we need for our equation.
Now that we have the appropriate values for a and b, let's apply the binomial theorem:
2Step 2: Apply the binomial theorem
Using the binomial theorem, we have:
\((a + b)^n = (1 + 2)^n = 3^n\)
Now, expanding the left hand side using the binomial theorem:
\((1+2)^n\ =\ \sum_{r=0}^{n} \binom{n}{r} 1^{n-r}2^{r}\)
But the powers of 1 have no effect on the sum, so we can simplify the expression:
3Step 3: Simplify the expression
Now, we can simplify the equation as:
\[\sum_{r=0}^{n} \binom{n}{r} 2^r\ =\ 3^n\]
This matches the expression we needed to prove, so now we can conclude that our proof is done.
4Step 4: Conclusion
By using the binomial theorem with appropriate values for a and b, we have proven that \(\sum_{r=0}^{n} 2^{r}\binom{n}{r} = 3^{n}\).
Key Concepts
CombinatoricsMathematical ProofDiscrete Mathematics
Combinatorics
Combinatorics is a fascinating branch of mathematics that focuses on counting, arrangement, and combination of objects. It's incredibly useful when solving problems that involve discrete and finite elements. In the context of the binomial theorem, combinatorics is the key to understanding how different values are selected and combined.
One of the most vital concepts in combinatorics is the binomial coefficient, denoted as \( \binom{n}{r} \). This represents the number of ways to choose \( r \) elements from a set of \( n \) elements, regardless of the order. It's expressed as a fraction:
One of the most vital concepts in combinatorics is the binomial coefficient, denoted as \( \binom{n}{r} \). This represents the number of ways to choose \( r \) elements from a set of \( n \) elements, regardless of the order. It's expressed as a fraction:
- \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \(!\) is the factorial function.
Mathematical Proof
Mathematical proofs are the backbone of ensuring statements in mathematics are true. Proofs bring rigor and logic, providing a solid foundation for understanding complex concepts. Here, we focus on proving identities using the binomial theorem.
In the provided exercise, mathematical proof is used to validate the equation:
In the provided exercise, mathematical proof is used to validate the equation:
- \( \sum_{r=0}^{n} 2^{r} \binom{n}{r} = 3^{n} \)
- \((1 + 2)^n = 3^n\)
- Expanded as \((1+2)^n\ =\ \sum_{r=0}^{n} \binom{n}{r} 1^{n-r}2^{r}\)
- \( \sum_{r=0}^{n} \binom{n}{r} 2^r\ =\ 3^n\)
Discrete Mathematics
Discrete mathematics is the study of structures that are distinct and separate (as opposed to continuous). It covers topics like graphs, integers, and yes, combinatorics. The finite nature of discrete mathematics makes it essential for computer science, algorithms, and encryption.
In this exercise, we're dealing with discrete quantities—integer values and fixed powers. The binomial theorem is a discrete mathematical tool that helps us expand powers of binomials into sums involving discrete amounts. For example:
In this exercise, we're dealing with discrete quantities—integer values and fixed powers. The binomial theorem is a discrete mathematical tool that helps us expand powers of binomials into sums involving discrete amounts. For example:
- The finite series \( \sum_{r=0}^{n} 2^{r}\) involves fixed powers and an integer coefficient \( \binom{n}{r} \).
- The expression \(3^n\) represents a distinct, singular quantity.
Other exercises in this chapter
Problem 30
A zip code in Canada consists of three letters and three digits. Each zip code begins with a letter. The letters and digits alternate; for instance, A1B2C3. Fin
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Find the number of solutions to each equation. $$x_{1}+x_{2}+x_{3}=12, x_{1}, x_{2} \geq 5,1 \leq x_{3} \leq 4$$
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Using the recursive definition of \(c_{n},\) compute each. $$c_{8}$$
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Find the number of ways a committee of five can be formed from a group of five boys and four girls, if each committee must contain: Exactly two boys.
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