To prove by mathematical induction that $1 + 4 + 7 + \dots + (3n-2) = \frac{n(3n-1)}{2}$ for all positive integers $n$. Let $P(n)$ be the statement $1 + 4 + 7 + \dots + (3n-2) = \frac{n(3n-1)}{2}$. Step 1: Base Case ($n=1$) Check if $P(1)$ is true. Left Hand Side (LHS): The sum of the first term is $1$. Right Hand Side (RHS): Substitute $n=1$ into the formula: $$ \frac{1(3(1)-1)}{2} = \frac{1(3-1)}{2} = \frac{1(2)}{2} = 1 $$ Since LHS = RHS, $P(1)$ is true. Step 2: Inductive Hypothesis Assume that $P(k)$ is true for some positive integer $k$. That is, assume: $$ 1 + 4 + 7 + \dots + (3k-2) = \frac{k(3k-1)}{2} $$ Step 3: Inductive Step (Prove $P(k+1)$) We need to show that $P(k+1)$ is true, i.e., $$ 1 + 4 + 7 + \dots + (3k-2) + (3(k+1)-2) = \frac{(k+1)(3(k+1)-1)}{2} $$ Consider the LHS of $P(k+1)$: $$ 1 + 4 + 7 + \dots + (3k-2) + (3(k+1)-2) $$ Using the Inductive Hypothesis, we can replace the sum up to $(3k-2)$: $$ \left( \frac{k(3k-1)}{2} \right) + (3(k+1)-2) $$ Simplify the last term: $$ \frac{k(3k-1)}{2} + (3k+3-2) $$ $$ \frac{k(3k-1)}{2} + (3k+1) $$ Combine the terms by finding a common denominator: $$ \frac{k(3k-1) + 2(3k+1)}{2} $$ Expand the numerator: $$ \frac{3k^2 - k + 6k + 2}{2} $$ $$ \frac{3k^2 + 5k + 2}{2} $$ Now, consider the RHS of $P(k+1)$: $$ \frac{(k+1)(3(k+1)-1)}{2} $$ Simplify inside the parenthesis: $$ \frac{(k+1)(3k+3-1)}{2} $$ $$ \frac{(k+1)(3k+2)}{2} $$ Expand the numerator: $$ \frac{k(3k) + k(2) + 1(3k) + 1(2)}{2} $$ $$ \frac{3k^2 + 2k + 3k + 2}{2} $$ $$ \frac{3k^2 + 5k + 2}{2} $$ Since the LHS equals the RHS, $P(k+1)$ is true. Conclusion: By the principle of mathematical induction, the statement $1 + 4 + 7 + \dots + (3n-2) = \frac{n(3n-1)}{2}$ is true for all positive integers $n$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.