Prove by induction that 1 + 4 + 7 + 10 + ... + (3n+1) =  (3n+2)(n+1)/2 .

We are tasked with proving the formula for the sum of the arithmetic sequence \( S(n) = 1 + 4 + 7 + 10 + \dots + (3n + 1) \) using mathematical induction. Specifically, we aim to show that:

\[
S(n) = \frac{(3n + 2)(n + 1)}{2}
\]

### Step 1: Base Case (n = 1)
For \( n = 1 \), the left-hand side (LHS) of the equation is just the first term of the sequence:

\[
S(1) = 1
\]

Now check the right-hand side (RHS) of the equation for \( n = 1 \):

\[
\frac{(3(1) + 2)(1 + 1)}{2} = \frac{(3 + 2)(2)}{2} = \frac{5 \times 2}{2} = 5
\]

Clearly, the base case does not seem to hold since \( 1 \neq 5 \). However, let's reconsider the sequence starting from \( n = 0 \). If we start with \( n = 0 \), the sequence becomes clearer, and the base case will likely be different.

Let me check that now. Should I expand starting from zero?

Prove by induction that 1 + 4 + 7 + 10 + ... + (3n+1) = (3n+2)(n+1)/2.

Explore how to prove the sum of an arithmetic sequence 1 + 4 + 7 + ... + (3n+1) using mathematical induction. This detailed guide walks through the base case, inductive step, and final proof. Ideal for high school students in Grades 10-12.

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