lesson -1

Let \(P(n)\) be the proposition:

\(\displaystyle 1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}\)

✅ Step 1: Base Case

Let \(n = 1\)

Left-hand side (LHS):
\(\displaystyle (2 \cdot 1 – 1)^2 = 1^2 = 1\)

Right-hand side (RHS):
\(\displaystyle \frac{1(4 \cdot 1^2 – 1)}{3} = \frac{3}{3} = 1\)

✅ Since LHS = RHS, the base case holds.

✅ Step 2: Inductive Hypothesis

Assume \(P(k)\) is true for some \(k \geq 1\), i.e.,

\(\displaystyle 1 + 9 + 25 + \cdots + (2k – 1)^2 = \frac{k(4k^2 – 1)}{3}\)

✅ Step 3: Inductive Step

We now prove \(P(k + 1)\) is true.
That is, we want to show:

\(\displaystyle 1 + 9 + \cdots + (2k – 1)^2 + (2k + 1)^2 = \frac{(k + 1)\left(4(k + 1)^2 – 1\right)}{3}\)

Using the inductive hypothesis:

\(\displaystyle \frac{k(4k^2 – 1)}{3} + (2k + 1)^2\)

Now expand:

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

So the total becomes:

\(\displaystyle \frac{k(4k^2 – 1)}{3} + 4k^2 + 4k + 1\)

Convert to a common denominator:

\(\displaystyle \frac{k(4k^2 – 1) + 3(4k^2 + 4k + 1)}{3}\)

Expand both parts:

\(\displaystyle k(4k^2 – 1) = 4k^3 – k\)
\(\displaystyle 3(4k^2 + 4k + 1) = 12k^2 + 12k + 3\)

Add:

\(\displaystyle \frac{4k^3 + 12k^2 + 11k + 3}{3}\)

Now simplify \(P(k + 1)\):

\(\displaystyle (k + 1)^2 = k^2 + 2k + 1\)

\(\displaystyle 4(k + 1)^2 – 1 = 4k^2 + 8k + 3\)

Then,

\(\displaystyle (k + 1)(4k^2 + 8k + 3) = 4k^3 + 12k^2 + 11k + 3\)

Therefore,

\(\displaystyle \frac{(k + 1)(4(k + 1)^2 – 1)}{3} = \frac{4k^3 + 12k^2 + 11k + 3}{3}\)

✅ So, \(P(k + 1)\) is true.

✅ Step 4: Conclusion

Since the formula holds for \(n = 1\), and
\(P(k) \Rightarrow P(k + 1)\),
it follows by mathematical induction that:

\(\displaystyle \boxed{1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}}\)
holds for all \(n \geq 1\).

Would you like a version exported in Word or PDF with typeset math and spacing as seen in textbooks? I can generate that for you instantly.