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Chapter 18: Problem 406

Prove by mathematical induction that \(1+7+13+\ldots+(6 n-5)=n(3 n-2)\)

Short Answer

Expert verified
To prove by mathematical induction that the sum of the arithmetic sequence \(1 + 7 + 13 + \ldots + (6n - 5) = n(3n - 2)\), we first establish the base case for \(n = 1\), which holds true as \(1(3 \cdot 1 -2) = 1\). Next, we assume the formula is true for an arbitrary positive integer \(k\), and then demonstrate that it's also true for \(n = k + 1\). After simplification, we obtain the equation \(3k^2 + 4k + 1 = 3k^2 + 4k + 1\), confirming the formula holds for \(n = k + 1\). Thus, by mathematical induction, the given formula provides the sum of the arithmetic sequence for all positive integers \(n\).

Step by step solution

01

Proving the base case

Before we start the induction, we need to prove the base case, i.e., the given formula holds true for \(n = 1\). According to the given formula, the sum \(S_1\) when \(n = 1\) is: \[S_1 = 1(3 \cdot 1 - 2)\] \[S_1 = 1\] Now, we shall check if the formula holds true when \(n = 1\). The sum of the arithmetic sequence with only the first term should be: \[S_1 = 1 \] As both values of \(S_1\) are equal, the base case is proven to be true.
02

Inductive Hypothesis

Now let's assume that the given formula is true for some arbitrary positive integer \(k\): \[1+7+13...+(6k-5) = k(3k-2) \qquad (1) \] This is our inductive hypothesis.
03

Proving the formula for \(n = k + 1\)

Using the method of mathematical induction, we now need to prove that the given formula also holds true for \(n = k + 1\). The sum of the arithmetic sequence up to \(n = k+1\) will be: \[1+7+13...+(6k-5)+(6(k+1)-5) = (k+1)(3(k+1)-2) \] By adding the next term of the sequence, \((6(k+1)-5)\), to both sides of equation \((1)\): \[1+7+13...+(6k-5) + (6(k+1)-5) = k(3k-2) + (6(k+1)-5) \qquad (2) \] Since we want to show that both sides of equation \((2)\) are equal, let's now simplify the right side of the equation: \[k(3k-2) + (6(k+1)-5) = k(3k-2) + 6k + 1\] Now, expand and simplify: \[(k+1)(3k+1) = 3k^2 + k - 2k + 6k + 1 \] \[3k^2 + 4k + 1 = 3k^2 + 4k + 1\] Since both sides of equation \((2)\) are equal, the formula holds true for \(n = k + 1\).
04

Conclusion

Since the given formula holds true for the base case of \(n = 1\), and we've demonstrated that if it's true for \(n = k\), it's also true for \(n = k + 1\), it follows that the sum of the given arithmetic sequence can be described by the formula \(n(3n-2)\) for all positive integers \(n\), via the principle of mathematical induction.

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Key Concepts

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

Arithmetic Sequence
An arithmetic sequence is a series of numbers in which each term after the first is obtained by adding a constant difference to the previous term. This constant difference is known as the 'common difference'. In our exercise, the sequence starts with the number 1 and continues with numbers that are each 6 units larger than the previous one, such as 1, 7, 13, and so on.

To identify the general term in an arithmetic sequence, we can use the formula:
  • First term = a
  • Common difference = d
  • Term number = n
The n-th term (\(a_n\)) is given by:\[ a_n = a + (n-1)d \]In our case, the formula for the n-th term is:\[ 6n - 5 \]This formula helps us determine any term within the sequence without listing all the previous terms.
Base Case
The base case in mathematical induction is the starting point of the proof. It's where we show that the statement or formula we want to prove is true for the first value of the variable, usually for \(n = 1\).

In our problem:
The base case requires us to check if the formula \(n(3n-2)\) holds when \(n = 1\). Computing this gives:
  • \(S_1 = 1(3\cdot1 - 2) = 1\)
  • The sum of sequences up to the first term: \(1\)
Both calculated sums are equal, confirming our base case is true for \(n = 1\). This establishes the foundation on which we can build the inductive proof.
Inductive Hypothesis
Formulating the inductive hypothesis is a crucial step in mathematical induction. At this stage, we assume that the statement we aim to prove is true for some arbitrary integer \(k\).

For our example, this involves the assumption:
  • \(1 + 7 + 13 + \, ... \, + (6k - 5) = k(3k - 2)\)
This assumption doesn't mean we've proven it yet, but it provides the basis to show that if it's true for any integer \(k\), it's also true for \(k + 1\). This leap from \(k\) to \(k + 1\) is what solidifies the inductive step of the proof.
Formula Proof
The final step is to prove the formula for the next integer, which means showing it works for \(n = k + 1\). Armed with our inductive hypothesis, we explore:
  • Add the next term: \((6(k+1) - 5)\) to both sides of the hypothesis equation.
  • Simplify: \(k(3k-2) + 6k + 1 = (k+1)(3(k+1)-2)\)
When both sides of the equation are equal after simplification, it confirms that the formula holds for \(n = k + 1\).

This proof technique demonstrates that the formula \(n(3n-2)\) accurately describes the sum of the arithmetic sequence for all natural numbers \(n\). Once both the base case and the inductive step are validated, the formula is proven for all enthusiastically awaited integers.

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Most popular questions from this chapter

Prove that the sum of the cubes of the first and natural numbers is equal to \(\\{\mathrm{n}(\mathrm{n}+1) / 2\\}^{2}\)

Prove by mathematical Induction that the number of straight lines determined by \(\mathrm{n}>1\) points, no 3 on the same straight line, is \(1 / 2 \mathrm{n}(\mathrm{n}-1)\)

Using mathematical induction, prove the binomial formula \((a+x)^{n}=a^{n}+n a^{n-1} x+\\{n(n-1) / 2 !\\} a^{n-2} x^{2}+\ldots\) \(+[\\{\operatorname{nn}(n-1) \ldots(n-r+2)\\} /\\{(r-1) !\\}] a^{n-r+1} x^{r-1}+\ldots+x^{n}\) for positive integral values of \(\mathrm{n}\).

Prove by mathematical induction \(1^{2}+2^{2}+3^{2}+\ldots+\mathrm{n}^{2}=(1 / 6) \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\)

Prove: \(1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+n(n+1)[n(n+1)(n+2)] / 3\)
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