1. Prove that 1^2+ 2^2+ 3^2 + ⋯ + 𝑛^2 = 𝑛(𝑛+1)(2𝑛+1)÷6, ∀𝑛𝑛 ∈ ℕ by using the principle of

Question

mathematical induction

Answer 1

First, we will prove the base case for n = 1:

When n = 1, the left side of the equation is 1^2 = 1.

Plugging n = 1 into the right side of the equation, we get:

1(1 + 1)(2(1) + 1)/6 = 1(2)(3)/6 = 6/6 = 1.

Since the left side and the right side both equal 1 when n = 1, the base case is true.

Now, we assume that the equation holds for some positive integer k, where k ≥ 1:

1^2 + 2^2 + 3^2 + ... + k^2 = k(k + 1)(2k + 1)/6.

We will prove that the equation also holds for k + 1:

1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = (k + 1)[(k + 1) + 1][2(k + 1) + 1]/6.

Starting with the left side:

1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = k(k + 1)(2k + 1)/6 + (k + 1)^2.

Multiplying both terms by 6:

6[k(k + 1)(2k + 1)/6 + (k + 1)^2] = k(k + 1)(2k + 1) + 6(k + 1)^2.

Expanding the multiplication:

k(k + 1)(2k + 1) + 6(k + 1)^2 = 2k^3 + 3k^2 + k + 6k^2 + 12k + 6.

Combining like terms:

2k^3 + 9k^2 + 13k + 6.

Factoring out a common factor of 2:

2(k^3 + 4.5k^2 + 6.5k + 3).

Factoring the cubic term:

2(k + 1)(k + 2)(k + 3).

Now, we have:

2(k + 1)(k + 2)(k + 3).

We can rewrite the right side of the equation as:

(k + 1)((k + 1) + 1)(2(k + 1) + 1)/6.

This is equal to:

(k + 1)(k + 2)(2k + 3)/6.

Therefore, we have shown that:

1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = (k + 1)(k + 2)(2k + 3)/6.

By the principle of mathematical induction, the statement is true for all positive integers n.