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Chapter 12: Problem 5

Use mathematical induction to prove that the formula is true for all natural numbers \(n\) $$5+8+11+\cdots+(3 n+2)=\frac{n(3 n+7)}{2}$$

Short Answer

Expert verified
The formula is true for all natural numbers \( n \) by mathematical induction.

Step by step solution

01

Base Case

When using mathematical induction, you start with the base case. To prove the statement for \( n = 1 \), substitute \( n = 1 \) into both sides of the equation. The left-hand side is the sum with one term: \( 5 \). The right-hand side is \[ \frac{1(3 \times 1 + 7)}{2} = \frac{1 \times 10}{2} = 5. \] The base case holds true.
02

Inductive Hypothesis

Assume the formula is true for some arbitrary natural number \( k \). That is, assume \[ 5 + 8 + 11 + \cdots + (3k + 2) = \frac{k(3k + 7)}{2}. \] This assumption is the inductive hypothesis.
03

Inductive Step

Show that the formula holds for \( n = k+1 \). Consider \[ 5 + 8 + 11 + \cdots + (3k + 2) + [3(k+1) + 2]. \] Using the inductive hypothesis, the sum is equal to \[ \frac{k(3k + 7)}{2} + (3k + 3 + 2). \] Simplify the new term: \( 3(k+1) + 2 = 3k + 5 \). Therefore, the expression becomes \[ \frac{k(3k + 7)}{2} + 3k + 5. \]
04

Simplification

Combine the terms:\[ \frac{k(3k + 7)}{2} + \frac{2(3k + 5)}{2} = \frac{k(3k + 7) + 2(3k + 5)}{2}. \]Distribute and combine like terms:\[ k(3k + 7) = 3k^2 + 7k \] and \[ 2(3k + 5) = 6k + 10. \]Add these together:\[ 3k^2 + 7k + 6k + 10 = 3k^2 + 13k + 10. \]
05

Verify the Formula

Confirm that the resulting expression meets the original formula for \( k+1 \):\[ \frac{(k+1)(3(k+1) + 7)}{2} = \frac{(k+1)(3k + 3 + 7)}{2} = \frac{(k+1)(3k + 10)}{2}. \]Distribute:\[ (k+1)(3k + 10) = 3k^2 + 10k + 3k + 10 = 3k^2 + 13k + 10. \]Since the expressions are equal, the formula holds for \( n = k+1 \). by mathematical induction, the formula is true for all natural numbers \( n \).
06

Conclusion

By mathematical induction, since the base case and the inductive step have been verified, the equation \[ 5 + 8 + 11 + \cdots + (3n + 2) = \frac{n(3n + 7)}{2} \] holds for all natural numbers \( n \).

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Key Concepts

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

Base Case
The journey of mathematical induction begins with the base case. This initial step is crucial because it acts as the foundation for proving that our formula works for every natural number. To do this, we start by checking if the formula holds true for the smallest natural number, which is usually 1. In our exercise:
  • We substitute \(n = 1\) into both sides of the equation.
  • On the left, the sum with just one term is \(5\).
  • The right side becomes \(\frac{1(3 \times 1 + 7)}{2} = \frac{10}{2} = 5\).
Both sides equal, confirming the formula is correct for \(n = 1\). This step is important because it shows that the formula works at the starting point, setting the stage for the next steps of induction.
Inductive Hypothesis
After proving the base case, we move on to the inductive hypothesis. This step is like making a strategic assumption to help us prove the formula's validity further. In simple terms, we assume the formula is true for some arbitrary natural number, \(k\). This is a temporary assumption meant to bridge the base case and the next step. In our example:
  • We assume \(5 + 8 + 11 + \cdots + (3k + 2) = \frac{k(3k + 7)}{2}\) is true.
This assumption lets us use the assumed truth for \(k\) to investigate if the formula holds for \(k+1\). The inductive hypothesis is not the proof itself but a necessary assumption that guides the complete induction process.
Inductive Step
The inductive step is where the true magic of induction happens. It involves proving that if the statement holds for a specific natural number \(k\), it must also hold for its next number, \(k+1\). To achieve this:
  • We add the next term, \(3(k+1) + 2\), to both sides of the assumed equation.
  • Using our inductive hypothesis, we have the expression: \(\frac{k(3k + 7)}{2} + (3k + 5)\).
  • We simplify this to confirm that it matches the formula for \(n = k+1\).
This step verifies the continuity of the truth, ensuring that if the formula works for \(k\), it will also work for \(k+1\), thus potentially for all natural numbers.
Natural Numbers
To understand mathematical induction better, it's important to be clear about the realm of natural numbers. Natural numbers are the infinite collection of numbers starting from 1 and increasing without bounds. They're the basis for counting and are fundamental to arithmetic.
In this context:
  • The primary goal of induction is to show a statement holds true across this entire set.
  • We start with the smallest natural number in the base case.
  • We prove for a generic \(k\) and extend it to \(k+1\) through the inductive step.
By tackling each number in the sequence, mathematical induction leverages the inherent properties of natural numbers to validate statements that span their vast set. This makes it a powerful and essential tool in mathematics for proving assertions about every member of this endless sequence.

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Most popular questions from this chapter

Which is larger, \((100 !)^{101}\) or (101!) \(^{100}\) ? [ Hint: Try factoring the expressions. Do they have any common factors?]

Paying off a Debt Margarita borrows \(\$ 10,000\) from her uncle and agrees to repay it in monthly installments of \(\$ 200\). Her uncle charges \(0.5 \%\) interest per month on the balance. (a) Show that her balance \(A_{n}\) in the \(n\) th month is given recursively by \(A_{0}=10,000\) and $$ A_{n}=1.005 A_{n-1}-200 $$ (b) Find her balance after six months.

A couple can afford to make a monthly mortgage payment of \(\$ 650 .\) If the mortgage rate is \(9 \%\) and the couple intends to secure a 30 -year mortgage, how much can they borrow?

Amortizing a Mortgage When they bought their house, John and Mary took out a \(\$ 90,000\) mortgage at \(9 \%\) interest, repayable monthly over 30 years. Their payment is \(\$ 724.17\) per month (check this, using the formula in the text). The bank gave them an amortization schedule, which is a table showing how much of each payment is interest, how much goes toward the principal, and the remaining principal after each payment. The table below shows the first few entries in the amortization schedule. $$\begin{array}{|c|c|c|c|c|}\hline \begin{array}{c}\text { Payment } \\\\\text { number }\end{array} & \begin{array}{c}\text { Total } \\\\\text { payment }\end{array} & \begin{array}{c}\text { Interest } \\\\\text { payment }\end{array} & \begin{array}{c}\text { Principal } \\\\\text { payment }\end{array} & \begin{array}{c}\text { Remaining } \\\\\text { principal }\end{array} \\\\\hline 1 & 724.17 & 675.00 & 49.17 & 89,950.83 \\\2 & 724.17 & 674.63 & 49.54 & 89,901.29 \\\3 & 724.17 & 674.26 & 49.91 & 89,851.38 \\\4 & 724.17 & 673.89 & 50.28 & 89,801.10 \\\\\hline\end{array}$$ After 10 years they have made 120 payments and are wondering how much they still owe, but they have lost the amortization schedule. (a) How much do John and Mary still owe on their mortgage? [Hint: The remaining balance is the present value of the \(240 \text { remaining payments. }]\) (b) How much of their next payment is interest, and how much goes toward the principal? [Hint: since \(9 \% \div 12=0.75 \%,\) they must pay \(0.75 \%\) of the remaining principal in interest each month.]

Simplify using the Binomial Theorem. $$\frac{(x+h)^{3}-x^{3}}{h}$$
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