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Chapter 3: Problem 10

Beweisen Sie die allgemeine binomische Formel $$ (a+b)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) a^{k} b^{n-k} $$ f眉r \(n \in \mathbb{N}_{0}\) mittels vollst盲ndiger Induktion.

Short Answer

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Question: Prove the general binomial formula using the principle of mathematical induction. Answer: By using mathematical induction, we have shown that the general binomial formula holds for all \(n \in \mathbb{N}_0\): $$(a+b)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l}n\\k\end{array}\right) a^{k} b^{n-k}$$

Step by step solution

01

Induction Base: Prove the formula for n=0

When \(n=0\), the general binomial formula becomes: $$(a+b)^0 = \sum_{k=0}^{0}\left(\begin{array}{l}0\\k\end{array}\right) a^k b^{0-k}$$ Since \(0^0\) and \((a+b)^0\) are both equal to 1 and the sum only has one term (k = 0) with binomial coefficient \(\binom{0}{0} = 1\), we have: $$(a+b)^0 = 1 = \left(\begin{array}{l}0\\0\end{array}\right) a^0 b^{0} = 1$$ So, the formula has been proven for the base case \(n=0\).
02

Induction Assumption: Formula holds for arbitrary n=k

Let's assume the general binomial formula works for an arbitrary \(n=k \in \mathbb{N}_0\), that is: $$(a+b)^k = \sum_{m=0}^{k}\left(\begin{array}{l}k\\m\end{array}\right) a^m b^{k-m}$$
03

Induction Step: Prove the formula for n=k+1

We now want to show that the general binomial formula also works for \(n=k+1\). Let's begin by expanding \((a+b)^{k+1}\) using the binomial theorem: $$(a+b)^{k+1} = (a+b) \cdot (a+b)^k = (a+b) \cdot \sum_{m=0}^{k}\left(\begin{array}{l}k\\m\end{array}\right) a^m b^{k-m}$$ Now distribute the \((a+b)\) term through the sum: $$(a+b)^{k+1} = \sum_{m=0}^{k}\left(\begin{array}{l}k\\m\end{array}\right) a^m b^{k-m} (a+b)$$ $$ = \sum_{m=0}^{k}\left(\begin{array}{l}k\\m\end{array}\right) a^{m+1} b^{k-m} + \sum_{m=0}^{k}\left(\begin{array}{l}k\\m\end{array}\right) a^m b^{k+1-m}$$ Now we can rewrite the sums by shifting the indices: $$ = \sum_{m=1}^{k+1}\left(\begin{array}{l}k\\m-1\end{array}\right) a^m b^{k+1-m} + \sum_{m=0}^{k}\left(\begin{array}{l}k\\m\end{array}\right) a^m b^{k+1-m}$$ Combine the two sums: $$(a+b)^{k+1} = \sum_{m=0}^{k+1}\left[\left(\begin{array}{l}k\\m-1\end{array}\right) + \left(\begin{array}{l}k\\m\end{array}\right)\right] a^m b^{k+1-m}$$ Now, from the property of binomial coefficients, we know: $$\left(\begin{array}{l}k+1\\m\end{array}\right) = \left(\begin{array}{l}k\\m-1\end{array}\right) + \left(\begin{array}{l}k\\m\end{array}\right)$$ Substitute this into the previous equation: $$(a+b)^{k+1} = \sum_{m=0}^{k+1}\left(\begin{array}{l}k+1\\m\end{array}\right) a^m b^{k+1-m}$$
04

Conclusion

Since we have proven the general binomial formula for the base case \(n=0\) and have shown that if the formula holds for \(n=k\), it also holds for \(n=k+1\), we can conclude by mathematical induction that the general binomial formula holds for all \(n \in \mathbb{N}_0\): $$(a+b)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l}n\\k\end{array}\right) a^{k} b^{n-k}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mathematical Induction
Mathematical induction is a powerful proof technique used in mathematics to establish the truth of an infinite series of statements. It's akin to a row of dominoes, where knocking down the first one causes all the rest to fall. The process of induction involves two crucial steps:
  • Base Case: This first step involves proving that the statement holds when the variable (often denoted as \(n\)) is at its smallest value, typically 0 or 1.
  • Inductive Step: In this step, you assume the statement is true for some arbitrary case \(n = k\), called the "inductive hypothesis," and then use this assumption to prove that the statement is also true for \(n = k+1\).
In the context of proving the Binomial Theorem, we initiate with the base case when \(n = 0\), verify its validity, then hypothesize correctness for \(n = k\), and demonstrate how this leads to correctness for \(n = k+1\). Induction thereby confirms the formula's truth for all natural numbers.
Binomial Coefficients
Binomial coefficients are the centerpiece of the Binomial Theorem. These coefficients are represented as \(\binom{n}{k}\), read as "n choose k", which solve the number of ways to choose \(k\) elements from a set of \(n\) elements.
  • The binomial coefficient \(\binom{n}{k}\) is calculated using the factorial formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(!\) denotes a factorial, the product of all positive integers up to that number.
  • These coefficients appear in the expansion of \((a+b)^n\), reflecting how different terms based on combinations of terms \(a^k\) and \(b^{n-k}\) come together.
  • An important property used in proofs like mathematical induction is known as Pascal鈥檚 identity: \(\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}\), which aids in manipulating and simplifying expressions during algebraic transformations.
Understanding binomial coefficients is crucial because they embody the combinatorial aspect of the Binomial Theorem, helping to weave together terms in the expansion.
Proof Techniques
When approaching mathematical proofs, a variety of techniques are employed to validate theorems. With the Binomial Theorem, mathematical induction serves as the chosen method. Here are some insights into executing proofs effectively:
  • Comprehend the Statement: Fully understanding what needs to be proven is key. Break down the components and theorems involved in the problem.
  • Choose the Right Tool: Identify which proof technique best fits. For recursive or progressive proofs, induction is effective as seen with the Binomial Theorem.
  • Follow Logical Steps: Establish a clear sequence of logical steps. Begin with the base case, set the inductive step, and conclude the inductive hypothesis successfully leads to the next case.
  • Verify Each Step: Each step in a proof must be closely scrutinized to ensure no logical errors occur. This validates the whole induction process.
  • Communicate Clearly: Writing out the proof with clarity helps ensure both the prover and the audience can follow the argument from start to finish without ambiguity.
The use of proof techniques in mathematics requires practice, patience, and precision. These aids in building solid foundations for the understanding and application of complex mathematical concepts.

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Most popular questions from this chapter

Warum werden leere Summen gleich null, leere Produkte aber gleich eins gesetzt?

Die Zahlen \(a_{k}\) mit \(k \in \mathbb{N}\) seien beliebig aus \(\mathbb{R}\). Eine Summe der Form $$ T_{n}=\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right) $$ nennt man eine Teleskopsumme. Bestimmen Sie eine geschlossene Formel f眉r den Wert einer solchen Summe und beweisen Sie sie mit Indexverschiebungen sowie mittels vollst盲ndiger Induktion.

K枚nnen Angaben von Werten 眉ber \(100 \%\) sinnvoll\\} sein?

Bestimmen Sie die Summe aller nat眉rlichen \(\mathrm{Zah}\) len von eins bis tausend.

Seltener als mit dem Binomialkoeffizienten hat man es mit seiner Verallgemeinerung, dem Multinomialkoeffizienten zu tun. Dieser ist definiert als $$ \left(\begin{array}{c} n \\ \left\\{k_{1}, \ldots, k_{m}\right\\} \end{array}\right)=\frac{n !}{k_{1} ! k_{2} ! \ldots k_{m} !} $$ mit Zahlen \(k_{i} \in \mathbb{N}_{0}\), die zus盲tzlich die Bedingung $$ k_{1}+k_{2}+\ldots+k_{m}=n $$ erfullen. Im Fall \(m=2\) reduziert sich das mit \(k_{1}=k\) und \(k_{2}=n-k\) auf den bekannten Binomialkoeffizienten. 鈥濫chte" Multinomialkoeffizienten treten dann auf, wenn man ein Multinom, also eine Summe mit mehr als zwei Summanden potenziert: $$ \begin{aligned} &\left(a_{1}+a_{2}+\ldots+a_{m}\right)^{n} \\ &\quad=\sum_{k_{1}+\ldots+k_{m}=n}\left(\begin{array}{c} n \\ \left\\{k_{1}, \ldots, k_{m}\right\\} \end{array}\right) a_{1}^{k_{1}} a_{2}^{k_{2}} \ldots a_{m}^{k_{m 1}} \end{aligned} $$ Bestimmen Sie die Multinomialkoeffizienten f眉r \(n=2\) und \(m=3\) und ermitteln Sie damit ohne Ausmultiplizieren den Ausdruck \((a+b+c)^{2}\)
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