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Chapter 1: Problem 2

Prove that \(1^{3}+2^{3}+\cdots+n^{3}=\left[\frac{1}{2} n(n+1)\right]^{2}\) for all \(n \in \mathbb{N}\).

Short Answer

Expert verified
By proving both the base case and the inductive step, the equation \(1^{3}+2^{3}+\cdots+n^{3}=\left[\frac{1}{2} n(n+1)\right]^{2}\) is proven true for all natural numbers \(n\).

Step by step solution

01

Base Case

The base case is tested by replacing \(n\) with \(1\). Thus, we have \(1^3 = \left[\frac{1}{2}(1)(1+1)\right]^2\). This simplifies to \(1 = 1\), which is true. Therefore, the base case holds.
02

Inductive Step

Assume the statement is true for \(n = k\). This means that \(1^3 + 2^3 + \cdots + k^3 = \left[\frac{1}{2}k(k+1)\right]^2\). This assumption is known as the inductive hypothesis. Now consider the next natural number \(k+1\). The sum of cubes is \(1^3 + 2^3 + \cdots + k^3 + (k+1)^3\). Substituting the inductive hypothesis, this expression becomes \[\left[\frac{1}{2}k(k+1)\right]^2 + (k+1)^3\]. To simplify further, write the expression in a common format, for example \[\left(\frac{1}{2}k^2k + \frac{1}{2}k^2 + k^3 + 3k^2 + 3k + 1\right)\] which simplifies to \[\left(\frac{(k+1)(k+2)}{2}\right)^2\], the general form for \(n = k + 1\), verifying our inductive step.
03

Conclusion

Since the statement is true for \(n = 1\) and we have shown that if it's true for \(n = k\), then it's true for \(n = k+1\), by the principle of mathematical induction, the statement \(1^{3}+2^{3}+\cdots+n^{3}=\left[\frac{1}{2} n(n+1)\right]^{2}\) is true for all \(n \in \mathbb{N}\).

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Most popular questions from this chapter

Prove that a nonempty set \(T_{1}\) is finite if and only if there is a bijection from \(T_{1}\) onto a finite set \(T_{2}\)

Draw diagrams in the plane of the Cartesian products \(A \times B\) for the given sets \(A\) and \(B\). (a) \(A=\\{x \in \mathbb{R}: 1 \leq x \leq 2\) or \(3 \leq x \leq 4\\}, B=\\{x \in \mathbb{R}: x=1\) or \(x=2\\}\). (b) \(A=\\{1,2,3\\}, B=\\{x \in \mathbb{R}: 1 \leq x \leq 3\\}\).

Prove that \(n^{3}+(n+1)^{3}+(n+2)^{3}\) is divisible by 9 for all \(n \in \mathbb{N}\).

Determine the number of elements in \(\mathcal{P}(S)\), the collection of all subsets of \(S\), for each of the following sets: (a) \(S:=\\{1,2\\}\) (b) \(\quad S:=(1,2,3)\), (c) \(S:=(1,2,3,4\\}\). Be sure to include the empty set and the set \(S\) itself in \(\mathcal{P}(S)\).

Prove the Principle of Mathematical Induction \(1.2 .3\) (second version).
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