Problem 47 - Algebra and Trigonometry

Question

If \(n\) is a positive integer, show that \(\left(\begin{array}{l}n \\\ 0\end{array}\right)+\left(\begin{array}{l}n \\\ 1\end{array}\right)+\cdots+\left(\begin{array}{l}n \\\ n\end{array}\right)=2^{n}\) [Hint: \(2^{n}=(1+1)^{n} ;\) now use the Binomial Theorem.]

Step-by-Step Solution

Answer

\[\left(\begin{array}{c} n \ 0\end{array}\right) + \left(\begin{array}{c} n \ 1\end{array}\right) + \cdots + \left(\begin{array}{c} n \ n\end{array}\right) = 2^n\]

Step 1: Understand the problem statement

The goal is to show that the sum of binomial coefficients for a given positive integer \(n\) equals \(2^n\). This can be written as: \[\left(\begin{array}{c}n \ 0\end{array}\right) + \left(\begin{array}{c}n \ 1\end{array}\right) + \cdots + \left(\begin{array}{c}n \ n\end{array}\right) = 2^n\]

Step 2: Recall the Binomial Theorem

The Binomial Theorem states that: \[(x + y)^n = \sum_{k=0}^{n} \left(\begin{array}{c} n \ k\end{array}\right) x^{n-k} y^{k}\]

Step 3: Apply the Binomial Theorem with \(x = 1\) and \(y = 1\)

Substitute \(x = 1\) and \(y = 1\) into the Binomial Theorem: \[(1 + 1)^n = \sum_{k=0}^{n} \left(\begin{array}{c} n \ k\end{array}\right) 1^{n-k} 1^{k}\] Since any number raised to any power is 1, this simplifies to: \[(1 + 1)^n = \sum_{k=0}^{n} \left(\begin{array}{c} n \ k\end{array}\right)\]

Step 4: Simplify the equation

Simplify \((1 + 1)^n\) to get: \[2^n = \sum_{k=0}^{n} \left(\begin{array}{c} n \ k\end{array}\right)\]

Step 5: Conclusion

Since we have shown that the sum of the binomial coefficients for a given \(n\) equals \(2^n\), we have proven the original statement.

Key Concepts

binomial coefficients

Let's explore the concept of *binomial coefficients*. These coefficients appear in binomial expansions and are denoted as \(\binom{n}{k}\). A binomial coefficient represents the number of ways to choose \(k\) items from \(n\) items without regard to the order.

In mathematical terms, it is expressed as: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(n!\) and \(k!\) denote the factorial of \(n\) and \(k\) respectively. These coefficients are integral in various combinatorial problems and are essential in understanding the Binomial Theorem.

You would typically find binomial coefficients in the expansion of \((x + y)^n\), where the coefficients are aligned with the terms in the polynomial. This understanding helps us see the relationship between binomial coefficients and combinations.

positive integer

A *positive integer* is simply any whole number greater than zero, such as 1, 2, 3, and so on. Positive integers are used to count discrete objects and appear throughout mathematics.

In our context, the variable \(n\) represents a positive integer. It defines the power to which a binomial is raised. When we talk about the theorem \((1+1)^n\), \(n\) determines the number of terms we will sum in our series.

Recognizing \(n\) as a positive integer ensures that we deal with a finite set of terms, making the summation and combinatorial proofs straightforward.

summation

The principle of *summation* is fundamental in mathematics. It involves adding a sequence of numbers to find their total. In the context of our exercise, we are summing binomial coefficients: \[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \text{...} + \binom{n}{n} \]

According to the Binomial Theorem, this summation is equivalent to \(2^n\). We used the theorem by setting \(x = 1\) and \(y = 1\), leading to: \[ (1 + 1)^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \text{...} + \binom{n}{n} \]

Simplifying the left-hand side, we get \[ (1+1)^n = 2^n \]

Thus, the sum of the binomial coefficients for a given positive integer \(n\) equals \(2^n\). Summation is a powerful tool to aggregate terms and find meaningful results in mathematics.