Final answer:
To prove the equation using mathematical induction, we show it holds for the base case and then assume it holds for an arbitrary value, proving it holds for the next value. By following this process, we can prove the given equation for all integers n ≥ 1.
Step-by-step explanation:
To prove the equation ∑i=1n(2i-1)² = n(2n-1)(2n+1)/3 for all integers n ≥ 1 using mathematical induction, we need to show that it holds for the base case n = 1 and then assume it holds for some arbitrary value k, and prove it holds for k + 1.
For the base case, when n = 1, we have (2*1-1)² = 1 and n(2n-1)(2n+1)/3 = 1(2(1)-1)(2(1)+1)/3 = 1. The equation holds true for n = 1.
Next, assume the equation holds for k, so ∑i=1k(2i-1)² = k(2k-1)(2k+1)/3. We need to prove that it also holds for k + 1.
By adding (2(k+1)-1)² to both sides of the assumed equation, we get:
∑i=1k+1(2i-1)² = k(2k-1)(2k+1)/3 + (2(k+1)-1)²
Simplifying both sides, we get:
k(2k-1)(2k+1)/3 + (2(k+1)-1)² = (2k³+3k²+k)/3 + (4k²+4k+1+4k+3) = (2k³+6k²+6k+4)/3 = (k+1)(2k+1)(2k+3)/3
This completes the proof by mathematical induction.