Question

Prove by induction on n that ∑n k=1 k2 = n(n 1)(2n 1)/6.

Answer 1

We will prove this by mathematical induction. The formula for the sum of the squares of the first n natural numbers is n(n+1)(2n+1)/6.

Step-by-step explanation:

We will prove this by mathematical induction.

1. Base case: When n = 1, the left-hand side (LHS) is equal to 1² = 1, and the right-hand side (RHS) is equal to 1(1-1)(2(1)-1)/6 = 1(0)(1)/6 = 0/6 = 0. Thus, the equation holds true for the base case.
2. Inductive step: Assume that the equation holds for some positive integer k: ∑k k=1 k² = k(k+1)(2k+1)/6.
3. We need to prove that the equation holds for k+1. Using the assumption, we have: ∑(k+1) k=1 k² = [k(k+1)(2k+1)/6] + (k+1)².
4. Simplifying the right-hand side, we get: [(k+1)(2k²+k) + 6(k+1)²]/6.
5. Distributing and simplifying further: [(2k³+3k²+k) + (6k²+12k+6)]/6 = (2k³+9k²+13k+6)/6.
6. Factoring out a (k+1) from the numerator, we have: (k+1)(2k²+7k+6)/6.
7. Using the factorization (k+2)(k+3)/2, we can rewrite the equation as: (k+1)(k+2)(k+3)/6.
8. Finally, we substitute back k for n, and we have proved that the equation holds for n+1: ∑n+1 k=1 k² = (n+1)(n+2)(2n+3)/6.

By mathematical induction, we have proven that ∑n k=1 k² = n(n+1)(2n+1)/6.