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Chapter 6: Problem 234

Prove by induction that $$ 5^{2 n+1} \cdot 2^{n+2}+3^{n+2} \cdot 2^{2 n+1} $$ is divisible by \(19,\) for all \(n \geqslant 0\)

Short Answer

Expert verified
The expression is divisible by 19 for all \( n \geq 0 \).

Step by step solution

01

Understanding Mathematical Induction

Mathematical induction is a proof technique used to prove statements about integers. The process involves two main steps: the base case and the induction step. In the base case, we prove the statement for an initial integer (typically, this is for \( n = 0 \) or \( n = 1 \)). In the induction step, we assume the statement holds for some arbitrary integer \( n = k \) and then prove it for \( n = k + 1 \). This establishes the validity of the statement for all integers starting from the base case.
02

Base Case

To prove the statement, we begin with the base case of \( n = 0 \). Substitute \( n = 0 \) into the expression:\[5^{2(0)+1} \cdot 2^{0+2} + 3^{0+2} \cdot 2^{2(0)+1}\]This simplifies to:\[5^1 \cdot 2^2 + 3^2 \cdot 2^1 = 5 \cdot 4 + 9 \cdot 2 = 20 + 18 = 38\]Now, we check divisibility:\[38 \div 19 = 2\]which confirms that 38 is divisible by 19, proving the base case.
03

Induction Hypothesis

Assume the statement holds for some integer \( n = k \), meaning:\[5^{2k+1} \cdot 2^{k+2} + 3^{k+2} \cdot 2^{2k+1} \equiv 0 \mod 19\]
04

Inductive Step

We must prove the statement for \( n = k + 1 \). Substitute \( n = k+1 \) into the expression:\[5^{2(k+1)+1} \cdot 2^{(k+1)+2} + 3^{(k+1)+2} \cdot 2^{2(k+1)+1}\]Simplifying this, we get:\[5^{2k+3} \cdot 2^{k+3} + 3^{k+3} \cdot 2^{2k+3}\]Rewriting using the induction hypothesis:\[5^{2k+1} \cdot 5^2 \cdot 2^{k+2} \cdot 2 + 3^{k+2} \cdot 3 \cdot 2^{2k+1} \cdot 2^2 = 5^2 \cdot 2 \left(5^{2k+1} \cdot 2^{k+2}\right) + 3 \cdot 4 \left(3^{k+2} \cdot 2^{2k+1}\right) \]Given that \( 5^{2k+1} \cdot 2^{k+2} + 3^{k+2} \cdot 2^{2k+1} \equiv 0 \mod 19 \) (from hypothesis), every term inside the multiples of 19 remains divisible by 19. Thus, the entire expression:\[5^{2k+3} \cdot 2^{k+3} + 3^{k+3} \cdot 2^{2k+3} \equiv 0 \mod 19\]proves the induction step.
05

Conclusion

Since we've established both the base case and the induction step, by mathematical induction, the expression \( 5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1} \) is divisible by 19 for all \( n \geq 0 \).

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Key Concepts

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

Divisibility
Divisibility is a foundational concept in number theory that determines if one number can be evenly divided by another without a remainder. In simpler terms, a number \(a\) is divisible by a number \(b\), if there exists an integer \(c\) such that \(a = b \cdot c\). This concept is crucial when examining expressions to determine their divisibility properties and is often the goal of proof techniques such as mathematical induction.

In our exercise, the expression \(5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}\) needs to be shown as divisible by \(19\) for all integers \(n\geq0\). We proved this by checking whether the expression evaluates to a multiple of \(19\) through both the base case and the inductive step. This strategy ensures that no matter what value \(n\) takes, the result will always be evenly divisible by \(19\).

Understanding divisibility and how to prove it using mathematical techniques not only enhances your number theory skills but also your logical thinking and problem-solving abilities in mathematics.
Induction Hypothesis
The induction hypothesis is a critical part of proving statements by mathematical induction. It is essentially an assumption made during the process which establishes a basis for further proving a hypothesis for the next step.

In our exercise, the induction hypothesis assumes that for some integer \(k\), the expression \(5^{2k+1} \cdot 2^{k+2} + 3^{k+2} \cdot 2^{2k+1}\) is divisible by \(19\). This assumption is not about proving immediately but using it as a stepping stone to show the expression holds true for the next consecutive integer, \(k+1\). Without this assumption, the structure of mathematical induction could not transition smoothly from one integer case to the next.

It鈥檚 crucial to understand the induction hypothesis, as this 'bridge' allows you to logically extend the truth of the original statement from one specific case to all larger cases.
Base Case
The base case is the starting point in the process of mathematical induction. It involves proving that a given statement, formula, or formula is true for the smallest possible integer, typically \(n = 0\) or \(n = 1\).

In our instance, we started with \(n = 0\) and evaluated the expression:
  • Substitute \(n = 0\) into \(5^{2(0)+1} \cdot 2^{0+2} + 3^{0+2}\cdot 2^{2(0)+1}\)
  • We simplified it to \(5 \cdot 4 + 9 \cdot 2 = 38\)
  • Since \(38 \div 19 = 2\) with no remainder, the base case holds
The base case confirms that the proposed statement is true for at least one integer, thereby providing the initial step necessary to apply the induction hypothesis and move forward with the proof.
Inductive Step
The inductive step is where the magic of mathematical induction truly happens. It's here that you extend the assumption made in the induction hypothesis to prove the next case.

Given the induction hypothesis that the statement is true for \(n = k\), the inductive step involves proving it for \(n = k + 1\). Using algebraic manipulation, we substitute \(n = k + 1\) into the expression:
  • Rewrite and simplify \(5^{2(k+1)+1} \cdot 2^{(k+1)+2} + 3^{(k+1)+2} \cdot 2^{2(k+1)+1}\).
  • With some factorization, leverage the induction hypothesis for \(n = k\).
  • Validate that the new expression remains divisible by \(19\).
By proving the inductive step, you demonstrate that if the statement holds for one integer (\(k\)), it must hold for the next (\(k+1\)), completing the logical chain that extends to all integers from the base case onward.

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Most popular questions from this chapter

Problem 236 Prove by induction the statement: $$ "(1+2+3+\cdots+n)^{2}=1^{3}+2^{3}+3^{3}+\cdots+n^{3}, \text { for all } n \geqslant 1 " $$ We now know that, for all \(n \geqslant 1\) : $$ 1+1+1+\cdots+1 \quad(n \text { terms })=n $$ And if we sum these "outputs" (that is, the first \(n\) natural numbers), we get the \(n^{\text {th }}\) triangular number: $$ 1+2+3+\cdots+n=\frac{n(n+1)}{2}=T_{n} $$ The next problem invites you to find the sum of these "outputs": that is, to find the sum of the first \(n\) triangular numbers.

(Calkin-Wilf tree) The binary tree in the plane has a distinguished vertex as 'root', and is constructed inductively. The root is joined to two new vertices; and each new vertex is then joined to two further new vertices \(-\) with the construction process continuing for ever (Figure 11 ). Label the vertices of the binary tree with positive fractions as follows: \- the root is given the label \(\frac{1}{1}\) \- whenever we know the label \(\frac{i}{j}\) of a 'parent' vertex, we label its 'left descendant' as \(\frac{i}{i+j},\) and its 'right descendant' \(\frac{i+j}{j}\). (a) Prove that every positive rational \(\frac{r}{s}\) occurs once and only once as a label, and that it occurs in its lowest terms. (b) Prove that the labels are left-right symmetric in the sense that labels in corresponding left and right positions are reciprocals of each other. \(\triangle\)

Interpret the following endless processes as infinite geometric series. (a) A square cake is cut into four quarters, with two perpendicular cuts through the centre, parallel to the sides. Three people receive one quarter each - leaving a smaller square piece of cake. This smaller piece is then cut in the same way into four quarters, and each person receives one (even smaller) piece - leaving an even smaller residual square piece, which is then cut in the same way. And so on for ever. What fraction of the original cake does each person receive as a result of this endless process? (b) I give you a whole cake. Half a minute later, you give me half the cake back. One quarter of a minute later, I return one quarter of the cake to you. One eighth of a minute later you return one eighth of the cake to me. And so on. Adding the successive time intervals, we see that $$ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots(\text { for ever })=1 $$ so the whole process is completed in exactly 1 minute. How much of the cake do I have at the end, and how much do you have?

(a) Experiment and guess a formula for the sum of the first \(n\) triangular numbers: $$ T_{1}+T_{2}+T_{3}+\cdots+T_{n}=1+3+6+\cdots+\frac{n(n+1)}{2} $$ (b) Prove by induction that your guessed formula is correct for all \(n \geqslant 1 . \triangle\) We now know closed formulae for $$ " 1+2+3+\cdots+n " $$ and for $$ " 1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+(n-1) n " $$ The next problem hints firstly that these identities are part of something more general, and secondly that these results allow us to find identities for the sum of the first \(n\) squares: $$ 1^{2}+2^{2}+3^{2}+\cdots+n^{2} $$ for the first \(n\) cubes: $$ 1^{3}+2^{3}+3^{3}+\cdots+n^{3} $$ and so on.

Prove by induction the statement: $$ " 1+3+5+\cdots+(2 n-1)=n^{2} \text { holds, for all } n \geqslant 1 " $$ The summation in Problem 231 was known to the ancient Greeks. The mystical Pythagorean tradition (which flourished in the centuries after Pythagoras) explored the character of integers through the 'spatial figures' which they formed. For example, if we arrange each successive integer as a new line of dots in the plane, then the sum " \(1+2+3+\cdots+n\) " can be seen to represent a triangular number. Similarly, if we arrange each odd number \(2 k-1\) in the sum " \(1+3+5+\cdots+(2 n-1)\) " as a " \(k\) -by- \(k\) reverse L-shape", or gnomon (a word which we still use to refer to the L-shaped piece that casts the shadow on a sundial), then the accumulated L-shapes build up an \(n\) by \(n\) square of dots \(-\) the "1" being the dot in the top left hand corner, the "3" being the reverse L-shape of 3 dots which make this initial "1" into a 2 by 2 square, the "5" being the reverse L-shape of 5 dots which makes this 2 by 2 square into a 3 by 3 square, and so on. Hence the sum " \(1+3+5+\cdots+(2 n-1)\) " can be seen to represent a square number. There is much to be said for such geometrical illustrations; but there is no escape from the fact that they hide behind an ellipsis (the three dots which we inserted in the sum between "5" and " \(2 n-1 "\), which were then summarised when arranging the reverse L-shapes by ending with the words "and so on"). Proof by mathematical induction, and its application in Problem 231, constitute a formal way of avoiding both the appeal to pictures, and the hidden ellipsis.
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