Respuesta :
Answer:
We will prove the given statement using the principle of mathematical induction.
First, we will establish the base case when n = 0:
∑²ₖ₀ (2k + 1)² = (0+1)(4(0)² + 8(0) + 3)/3
= 1(0+0+3)/3
= 3/3
= 1
So the statement holds true for n = 0.
Now, let's assume that the statement is true for some arbitrary positive integer k. That is,
∑²ₖ₀ (2k + 1)² = (k+1)(4k² + 8k + 3)/3
Now, we will prove that the statement holds true for k+1.
We need to prove that:
∑²ₖ₀ (2(k+1) + 1)² = ((k+1)+1)(4(k+1)² + 8(k+1) + 3)/3
Expanding the left side:
(2(k+1) + 1)² + ∑²ₖ₁ (2k + 1)²
= (2k + 3)² + ∑²ₖ₀ (2k + 1)²
= 4k² + 12k + 9 + ∑²ₖ₀ (2k + 1)²
= 4k² + 8k + 4k + 9 + ∑²ₖ₀ (2k + 1)²
= 4k² + 8k + 3 + (4k + 1) + ∑²ₖ₀ (2k + 1)²
= 4k² + 8k + 3 + (4k + 1) + (k+1)(4k² + 8k + 3 )/3 [Using our assumption]
= (k+2)(4k² + 8k + 3)/3 + (k+1)(4k² + 8k + 3 )/3
= ((k+2)(4k² + 8k + 3) + 3(k+1)(4k² + 8k + 3))/(3*1)
= (4k³ + 12k² + 9k + 8k² + 24k + 12 + 12k² + 36k + 27)/3
= (4k³ + 20k² + 72k + 27)/3
= ((k+2)(4(k+1)² + 8(k+1) + 3))/3
Thus, the statement holds true for k+1.
By the principle of mathematical induction, the statement holds true for all n ∈ N.
