Q. 28 In Problems 26鈥�28, use the Pri... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 12: Q. 28 (page 839)

In Problems 26鈥�28, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers.
$1^{2} + 4^{2} + 7^{2} + . . . . . . . {(3 n - 2)}^{2} = \frac{1}{2} n (6 n^{2} - 3 n - 1)$

Short Answer

Expert verified

Given statement is true for all natural numbers.

Step by step solution

Step 1. Given information.

Statement:

$1^{2} + 4^{2} + 7^{2} + . . . . . . . {(3 n - 2)}^{2} = \frac{1}{2} n (6 n^{2} - 3 n - 1)$

Step 2. Show that statement is true for n=1.

$L . H . S . = 1^{2} = 1$

$R . H . S . = \frac{1}{2} . 1 (6 . 1^{2} - 3.1 - 1) = \frac{1}{2} (6 - 3 - 1) = \frac{1}{2} (6 - 4) = \frac{2}{2} = 1$

Since $L . H . S . = R . H . S .$

Hence, statement is true for 1.

Step 3. Assume that statement is true for n=k.

$1^{2} + 4^{2} + 7^{2} + . . . . . . . {(3 k - 2)}^{2} = \frac{1}{2} k (6 k^{2} - 3 k - 1)$

Step 4. Show that the statement is true for n=k+1.

$L . H . S . = 1^{2} + 4^{2} + 7^{2} + . . . . . . . {(3 k - 2)}^{2} + (3 {(k + 1) - 2)}^{2} = \frac{1}{2} k (6 k^{2} - 3 k - 1) + {(3 k + 3 - 2)}^{2} = \frac{1}{2} k (6 k^{2} - 3 k - 1) + {(3 k + 1)}^{2} = \frac{1}{2} (6 k^{3} - 3 k^{2} - k) + (9 k^{2} + 1 + 18 k) = \frac{(6 k^{3} - 3 k^{2} - k) + 2 (9 k^{2} + 1 + 18 k)}{2} = \frac{(6 k^{3} - 3 k^{2} - k + 18 k^{2} + 2 + 36 k)}{2} = \frac{6 k^{3} + 15 k^{2} + 35 k + 2}{2}$ Use result of step 3.

$R . H . S . = \frac{1}{2} (k + 1) (6 {(k + 1)}^{2} - 3 (k + 1) - 1) = \frac{1}{2} (k + 1) (6 (k^{2} + 2 k + 1) - 3 k - 3 - 1) = \frac{1}{2} (k + 1) (6 k^{2} + 12 k + 6 - 3 k - 4) = \frac{1}{2} (k + 1) (6 k^{2} + 9 k + 2) = \frac{6 k^{3} + 15 k^{2} + 35 k + 2}{2}$

Since $L . H . S . = R . H . S .$ for $n = k + 1$

Hence the statement is true for all natural number

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91影视

In Problems 26鈥�28, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers.
$1^{2} + 4^{2} + 7^{2} + . . . . . . . {(3 n - 2)}^{2} = \frac{1}{2} n (6 n^{2} - 3 n - 1)$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Show that statement is true for n=1.

Step 3. Assume that statement is true for n=k.

Step 4. Show that the statement is true for n=k+1.

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91影视

In Problems 26鈥�28, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers.12+42+72+.......(3n-2)2=12n(6n2-3n-1)

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Show that statement is true for n=1.

Step 3. Assume that statement is true for n=k.

Step 4. Show that the statement is true for n=k+1.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Pure Maths

Theoretical and Mathematical Physics

Statistics

Probability and Statistics

Calculus

Study anywhere. Anytime. Across all devices.

In Problems 26鈥�28, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers.
$1^{2} + 4^{2} + 7^{2} + . . . . . . . {(3 n - 2)}^{2} = \frac{1}{2} n (6 n^{2} - 3 n - 1)$