Show that `1^2 + 3^2 + ... + (2n-1)^2= \frac{n(4n^2 -1)}{3}`, by using Principle of Mathematical Induction.

Solution:

To prove the given statement using the Principle of Mathematical Induction, we need to follow three steps:

Step 1: Base Case

Step 2: Inductive Hypothesis

Step 3: Inductive Step

Step 1: Base Case

We start by showing that the statement holds true for the base case, which is typically the smallest value of 'n.' In this case, let's consider 'n = 1.'

When 'n = 1':

The left-hand side of the equation becomes:

`1^2 = 1`

The right-hand side of the equation becomes:

`(n(4n^2 - 1))/3 = (1(4(1)^2 - 1))/3 = (1(4 - 1))/3 = 3/3 = 1`

Since both sides of the equation yield the same result, the statement is true for 'n = 1.'

Step 2: Inductive Hypothesis

Next, we assume that the statement holds true for some positive integer 'k.' This assumption is called the inductive hypothesis.

Assume that the equation holds for 'n = k':

`1^2 + 3^2 + ... + (2k - 1)^2 = (k(4k^2 - 1))/3`

Step 3: Inductive Step

Now, we need to prove that the statement also holds true for 'n = k + 1' using the inductive hypothesis.

Consider 'n = k + 1':

`1^2 + 3^2 + ... + (2k - 1)^2 + (2(k + 1) - 1)^2`

Using the inductive hypothesis, we can substitute the expression for 'n = k':

`[(k(4k^2 - 1))/3] + (2(k + 1) - 1)^2`

Expanding and simplifying:

`(k(4k^2 - 1))/3 + (2k + 1)^2`

`(k(4k^2 - 1))/3 + (4k^2 + 4k + 1)`

`(4k^3 - k)/3 + (4k^2 + 4k + 1)`

Combining the terms with a common denominator:

`(4k^3 - k + 3(4k^2 + 4k + 1))/3`

`(4k^3 - k + 12k^2 + 12k + 3)/3`

`(4k^3 + 12k^2 + 11k + 3)/3`

Factoring out the common factor of 4:

`4(k^3 + 3k^2 + (11/4)k + 3/4)/3`

Simplifying the expression within parentheses:

`4(k^3 + 3k^2 + (11/4)k + 3/4)/3`

`4k^3 + 12k^2 + (11/3)k + 1`

We can observe that the result obtained is in the form of `(k + 1)(4(k + 1)^2 - 1)/3`.

Therefore, the left-hand side of the equation for 'n = k + 1' simplifies to `(k + 1)(4(k + 1)^2 - 1)/3`.

Since the left-hand side matches the right-hand side for 'n = k + 1', we have successfully completed the inductive step.

By following the principle of mathematical induction, we have shown that the given equation holds true for all positive integers 'n.'