Chapter 12: Q. 28 (page 839)

In Problems 26–28, 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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In Problems 26–28, 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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In Problems 26–28, 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

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