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Proof (by mathematical induction): Let P(n) be the equation 1 + 6 + 11 + 16 + ? + (5n ? 4) = n(5n ? 3) / 2 We will show that P(n) is true for every integer n ? 1. Show that P(1) is true: Select P(1) from the choices below. 1 = 1 · (5 · 1 ? 3) / 2 The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer k ? 1, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ? 1, and suppose that P(k) is true. The left-hand side of P(k) is 1 + 6 + 11 + 16 + ? + (5k ? 4) and the right-hand side of P(k) is [The inductive hypothesis states that the two sides of P(k) are equal.] We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + ? + (5(k + 1) ? 4) = . After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes k(5k ? 3)/2 + (5(k + 1) ? 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal . Hence P(k + 1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]

          Proof (by mathematical induction): Let P(n) be the equation
1 + 6 + 11 + 16 + ? + (5n ? 4) = n(5n ? 3) / 2
We will show that P(n) is true for every integer n ? 1.
Show that P(1) is true: Select P(1) from the choices below.
1 = 1 · (5 · 1 ? 3) / 2
The selected statement is true because both sides of the equation equal the same quantity.
Show that for each integer k ? 1, if P(k) is true, then P(k + 1) is true:
Let k be any integer with k ? 1, and suppose that P(k) is true. The left-hand side of P(k) is 1 + 6 + 11 + 16 + ? + (5k ? 4) and the right-hand side of P(k) is
[The inductive hypothesis states that the two sides of P(k) are equal.]
We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + ? + (5(k + 1) ? 4) = . After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes k(5k ? 3)/2 + (5(k + 1) ? 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal . Hence P(k + 1) is true, which completes the inductive step.
[Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]

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Proof (by mathematical induction): Let P(n) be the equation
1 + 6 + 11 + 16 + ? + (5n ? 4) = n(5n ? 3) / 2
We will show that P(n) is true for every integer n ? 1.
Show that P(1) is true: Select P(1) from the choices below.
1 = 1 · (5 · 1 ? 3) / 2
The selected statement is true because both sides of the equation equal the same quantity.
Show that for each integer k ? 1, if P(k) is true, then P(k + 1) is true:
Let k be any integer with k ? 1, and suppose that P(k) is true. The left-hand side of P(k) is 1 + 6 + 11 + 16 + ? + (5k ? 4) and the right-hand side of P(k) is
[The inductive hypothesis states that the two sides of P(k) are equal.]
We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + ? + (5(k + 1) ? 4) = . After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes k(5k ? 3)/2 + (5(k + 1) ? 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal . Hence P(k + 1) is true, which completes the inductive step.
[Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]

Added by Darlene O.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ben Blakesley

10/14/2022

Step 1

.. + (5n - 1). This is an arithmetic series where the common difference is 5 and the first term is 1. Show more…

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Proof (by mathematical induction): Let P(n) be the equation 1 + 6 + 11 + 16 + ... + (5n - 4) = n(5n - 3) / 2 We will show that P(n) is true for every integer n ≥ 1. Show that P(1) is true: Select P(1) from the choices below. 1 = 1 · (5 · 1 - 3) / 2 The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer k ≥ 1, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ≥ 1, and suppose that P(k) is true. The left-hand side of P(k) is 1 + 6 + 11 + 16 + ... + (5k - 4) and the right-hand side of P(k) is [The inductive hypothesis states that the two sides of P(k) are equal.] We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + ... + (5(k + 1) - 4) = . After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes k(5k - 3)/2 + (5(k + 1) - 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal . Hence P(k + 1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]

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Transcript

00:01 Okay, so we want to show that this is true by induction.

00:04 So p of n is this equation.

00:06 So the first thing we want to do is show that p of 1 is true.

00:09 So if we sub in n equals 1, so the first thing is p of 1.

00:13 If we sub in n equals 1, well, the left -hand side is just 1, and the right -hand side is 1 times 5 times 1 minus 3 over 2.

00:23 And you can see that this is clearly correct because 5 times 1 is 5.

00:27 5 minus 3 is 2, 2 times 1 is 2, 2.

00:31 And 2 divided by 2 is 1.

00:33 So, p of 1 is correct.

00:35 So the next thing we want to do, so this is p of 1.

00:38 The next thing we want to do is assume that it's true for k, and then show that this implies that p of k plus 1 is true.

00:46 Okay, so we're going to let k be some integer, and we suppose that p of k is true.

00:53 So this means that the left -hand side of p of k is just 1 plus 6 plus 11 plus 16 plus dot, dot, dot, dot.

01:02 Plus 5n minus, not 5n of course, 5k minus 4.

01:08 This is the left -hand side of p of k, and the right -hand side is k times 5k minus 3 over 2.

01:19 So we're assuming that this is equal, that both of these sides are equal.

01:27 Okay, so what do we do next? so this is the right -hand side.

01:32 So this is this step.

01:34 So the inductive hypothesis states that the two sides of p of k equal.

01:38 So we know that 1 plus 6 plus 11 plus 16 plus dot dot dot plus 5k minus 4 is equal to k times 5k minus 3 over 2.

01:52 So this is what we know.

01:53 Now we have to show that p of k plus 1 is true.

01:58 Well p of k plus 1 is the following equation.

02:00 It's just 1 plus 6 plus 11 plus 16 plus 16 plus 1.

02:04 Plus dot dot plus 5 times k plus 1 minus 4.

02:12 And the right hand side is just what you get when you plug in k plus 1.

02:15 This is going to be k plus 1 times 5 k plus 1 minus 3 over 2...

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