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

Question

unicwaan · Answer

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

mathstudent55 · Answer

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

mathstudent55 · Answer

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?

mathstudent55 · Answer

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

mathstudent55 · Answer

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} $

mathstudent55 · Answer

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.

madhu.mukherjee.946 · Answer

@mathstudent55 superb

mathstudent55 · Answer

@madhu.mukherjee.946
Thanks!

unicwaan · Answer

Thank you for showing me the walk-through steps! Sorry that I went away from my compute! @mathstudent55