How is the inductive step proved for the conjecture $2 + 4$ + ... + $2n = n(n+1)$?

Conditions

• Inductive hypothesis: $P(k)$ is true.
• Goal: Prove $P(k+1)$ is true.
• Algebraic manipulation involves factoring common terms.

Reasoning, step by step

1. State the inductive hypothesis: $2 + 4$ + ... + $2k = k(k+1)$.
2. Write the LHS for $P(k+1)$: $2 + 4$ + ... + $2k + 2(k+1)$.
3. Substitute the hypothesis: Replace $2 + 4$ + ... + 2k with $k(k+1)$.
4. Simplify: $k(k+1) + 2(k+1) = (k+1)(k+2)$.
5. Calculate the RHS for $P(k+1)$: Substitute $n=k+1$ into $n(n+1)$ to get ($k+1$)($k+1+1$) = ($k+1$)($k+2$).
6. Compare LHS and RHS: They are identical.
7. Conclude $P(k)$ implies $P(k+1)$.

Example

The video derives: '$P(k+1)$_LHS = $2 + 4$ + ... + $2k + 2(k+1) = k(k+1) + 2(k+1) = (k+1)(k+2)$.' It then shows '$P(k+1)$_RHS = ($k+1$)($k+1+1$) = ($k+1$)($k+2$).' The narrator confirms, 'The clip therefore shows that the assumed truth of $P(k)$ leads to the truth of $P(k+1)$.'

Common misconceptions

• Forgetting to add the new term $2(k+1)$ to the LHS.
• Incorrectly factoring $k(k+1) + 2(k+1)$.
• Confusing the RHS of $P(k)$ with the RHS of $P(k+1)$.

Watch the explanation

Explore next

• Why can $k(k+1) + 2(k+1)$ be rewritten as ($k+1$)($k+2$)?
• What is the final conclusion of the induction proof?