Use Mathematical Induction to show that the statement

2 + 6 + 10 + . . . + (4n – 2) = 2n^2
is true

Using the 3 step process:

1. test for n=1
LS = 2
RS = 2(1^2) = 2

2. assume it true for n=k
that is ....
2+4+6 + ... + (4k-2) = 2k^2

3 . prove it is then true for =k+1
or
2+4+6+ ... + (4k-2) + 4(k+1)-2 = 2(k+1)2
LS = [2+4+6+...+ 4k-2 ] + 4(k+1)-2
= 2k^2 + 4k + 4 - 2
= 2(k^2 + 2k + 1)
= 2(k+1)^2
= RS

QED

Well, well, well! It looks like we have a mathematical induction problem at hand.

First, let's start by checking the base case. When n = 1, we have:

2 = 2(1)^2

2 = 2

Looking good so far! Now, let's assume that the statement holds true for some arbitrary k. That is, let's assume:

2 + 6 + 10 + . . . + (4k – 2) = 2k^2

Now, we need to prove that the statement also holds true for k+1. So let's add (4k + 2) to both sides of the equation:

2 + 6 + 10 + . . . + (4k – 2) + (4k + 2) = 2k^2 + (4k + 2)

Simplifying the left side gives us:

2 + 6 + 10 + . . . + (4k – 2) + (4k + 2) = 2k^2 + 4k + 2

Now, let's manipulate the right side:

2k^2 + 4k + 2 = 2(k^2 + 2k + 1)

2k^2 + 4k + 2 = 2(k + 1)^2

Ah, fantastic! We now have the same expression on both sides of the equation. This means that if the statement holds true for k, it also holds true for k+1!

And with that, we have successfully completed the mathematical induction. The statement

2 + 6 + 10 + . . . + (4n – 2) = 2n^2

is true for all positive integers n.

Mathematical induction, turning skeptics into believers since who knows when!

To prove the statement using mathematical induction, we need to follow these steps:

Step 1: Base Case
We first show that the statement is true for the initial value of n. Let's consider n = 1.
When n = 1,
2 + 6 + 10 + . . . + (4 × 1 – 2) = 2 × 1^2
2 + 6 + 10 + . . . + 2 = 2
2 = 2

Hence, the statement is true for n = 1.

Step 2: Inductive Hypothesis
Assume that the statement is true for some positive integer k, where k ≥ 1.
That means,
2 + 6 + 10 + . . . + (4k – 2) = 2k^2 --- [1]

Step 3: Inductive Step
We need to prove that the statement is also true for k + 1.
That means,
2 + 6 + 10 + . . . + (4(k + 1) – 2) = 2(k + 1)^2
We need to manipulate the left-hand side of the equation to look like the right-hand side, using the assumption we made in the inductive hypothesis.

Using [1], we have:
2 + 6 + 10 + . . . + (4k – 2) + (4(k + 1) – 2) = 2k^2 + (4(k + 1) – 2)
Simplifying,
2 + 6 + 10 + . . . + (4k – 2) + (4k + 4 – 2) = 2k^2 + 4(k + 1) – 2
2 + 6 + 10 + . . . + (4k – 2) + 4k + 2 = 2k^2 + 4(k + 1) – 2
(2 + 6 + 10 + . . . + (4k – 2)) + 4k + 2 = 2k^2 + 4k + 4 – 2

Now, substitute the value of the sum of the first k terms using the inductive hypothesis:
2k^2 + 4k + 2 = 2k^2 + 4k + 2

Hence, the statement is true for k + 1.

Step 4: Conclusion
By mathematical induction, the statement is true for all positive integers n.

Therefore, we have proven that 2 + 6 + 10 + . . . + (4n – 2) = 2n^2 for all positive integers n.

n an sum n

1 2 2
2 6 8
3 10 18
4 14 32

n (4n-2) 2n^2

(n+1) (4(n+1)-2) [2n^2+ (4(n+1)-2)] or hopefully 2(n+1)^2

[2n^2+ (4(n+1)-2)] = 2n^2+4n +2
and
2(n+1)^2 = 2(n^2+2n+1) = 2n^2+4n+2 done

2n^2+ (4(n+1)-2)