## unicwaan one year ago Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

1. unicwaan

$1^2+4^2+7^2+...+(3n−2)^2=\frac{ n(6n2−3n−1) }{ 2 }$

2. mathstudent55

An induction proof works like this: 1. Show the formula works for n = 1 2. Assume it works for n = k 3. Show it works for n = k + 1

3. mathstudent55

Let's start. Can you show the formula works for n = 1? Copy the formula. Then rewrite the formula replacing n with 1. What do you get?

4. mathstudent55

$$\Large 1^2+4^2+7^2+...+(3n−2)^2=\dfrac{ n(6n2−3n−1) }{ 2 }$$ Let n = 1 $$\Large [3(1)−2)]^2=\dfrac{ 1(6 \cdot 1^22−3 \cdot 1−1) }{ 2 }$$ $$\Large 1 = \dfrac{1(2)}{2}$$ $$\Large 1 = 1$$ We proved the expression works for n = 1.

5. mathstudent55

Now we assume the expression is true for n = k: $$\Large 1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = \dfrac{k(6k^2 - 3k - 1)}{2}$$

6. mathstudent55

Now we need o prove it works for n = k + 1 $$\Large 1^2 + 4^2 + 7^2 + ...+ (3k - 2)^2 + [3(k + 1) - 2]^2 =$$ $$\Large = \dfrac{(k + 1) [6(k + 1)^2 - 3(k + 1) - 1]}{2}$$ $$\Large = \dfrac{(k + 1)[6(k^2 + 2k + 1) - 3k -3 - 1]}{2}$$ $$\Large = \dfrac{(k + 1)(6k^2 + 12k + 6 - 3k - 4)}{2}$$ $$\Large = \dfrac{(k + 1)[6(k^2 + 2k + 1) - 3k - 3 - 1)}{2}$$ $$\Large = \dfrac{(k + 1)[6(k + 1)^2 - 3(k + 1) - 1)}{2}$$ The last expression above is the formula with k replaced by k + 1. This shows the formula works for n = k + 1. By induction, the formula has been proved to be true.

@mathstudent55 superb

8. mathstudent55