Prove by mathematical induction that: 1 + 7 + 13 + … + (6n-5) = n(3n-2)

To prove the statement \( S(n) \): 

\[
1 + 7 + 13 + \dots + (6n - 5) = n(3n - 2)
\]

by **mathematical induction**, we follow these steps:

### Step 1: Base Case (n = 1)

For \( n = 1 \), the left-hand side of the equation is just the first term of the sequence, which is:

\[
6(1) - 5 = 1
\]

The right-hand side of the equation for \( n = 1 \) is:

\[
1(3(1) - 2) = 1(3 - 2) = 1
\]

Since both sides are equal, the base case holds true.

### Step 2: Inductive Hypothesis

Assume that the statement holds true for \( n = k \), i.e.,

\[
1 + 7 + 13 + \dots + (6k - 5) = k(3k - 2)
\]

This is our **inductive hypothesis**.

### Step 3: Inductive Step

We need to prove that the statement holds for \( n = k + 1 \), i.e.,

\[
1 + 7 + 13 + \dots + (6(k + 1) - 5) = (k + 1)(3(k + 1) - 2)
\]

Start with the left-hand side:

\[
1 + 7 + 13 + \dots + (6k - 5) + (6(k + 1) - 5)
\]

Using the inductive hypothesis, we can replace the sum up to \( (6k - 5) \) by \( k(3k - 2) \), so the left-hand side becomes:

\[
k(3k - 2) + (6(k + 1) - 5)
\]

Simplify \( 6(k + 1) - 5 \):

\[
6(k + 1) - 5 = 6k + 6 - 5 = 6k + 1
\]

Thus, the left-hand side becomes:

\[
k(3k - 2) + (6k + 1)
\]

Expand both terms:

\[
3k^2 - 2k + 6k + 1 = 3k^2 + 4k + 1
\]

Now, simplify the right-hand side for \( n = k + 1 \):

\[
(k + 1)(3(k + 1) - 2) = (k + 1)(3k + 3 - 2) = (k + 1)(3k + 1)
\]

Expand:

\[
(k + 1)(3k + 1) = k(3k + 1) + 1(3k + 1) = 3k^2 + k + 3k + 1 = 3k^2 + 4k + 1
\]

Thus, the right-hand side equals \( 3k^2 + 4k + 1 \), which is the same as the left-hand side.

### Conclusion:

Since both sides are equal, the statement holds for \( n = k + 1 \). Therefore, by the principle of mathematical induction, the given statement is true for all \( n \geq 1 \).

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Would you like more details or have any questions?

Here are 5 related questions you can explore:

1. How does mathematical induction work in general?
2. Can you prove a different sum of arithmetic sequences by induction?
3. How do you derive a formula for the sum of the first \( n \) terms in an arithmetic progression?
4. What is the difference between weak and strong induction?
5. Can mathematical induction be applied to geometric sequences?

**Tip**: Always verify the base case carefully in induction, as it anchors the entire proof.

Learn how to prove the arithmetic sequence formula 1 + 7 + 13 + … + (6n-5) = n(3n-2) using mathematical induction. This step-by-step guide explains key algebraic concepts and the principle of mathematical induction. Suitable for high school students in Grades 10-12.

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