Final answer:
By expanding and simplifying the equation, we proved that the left side is equal to the right side, thus the formula is true for all natural numbers n.
Step-by-step explanation:
To prove that the formula is true for all natural numbers n using mathematical induction, we need to show that the formula is true for the base case (n = 1), and then assume that it is true for some arbitrary natural number k, and prove that it is also true for k + 1.
Base Case: When n = 1, the formula becomes: 2³ = 2*1²*(1+1)²
Simplifying both sides, we get: 8 = 2*1*4
This is true, so the formula holds for the base case.
Inductive Step: Assuming that the formula is true for some arbitrary natural number k, we have:
2³ + 4³ + 6³ + ... + (2k)³ = 2k²*(k+1)²
We need to prove that the formula also holds for k + 1:
2³ + 4³ + 6³ + ... + (2k)³ + (2(k+1))³ = 2(k+1)²*(k+1+1)²
To prove this, we start with the left side of the equation and use the assumption that the formula holds for k:
LHS = 2³ + 4³ + 6³ + ... + (2k)³ + (2(k+1))³
= 2k²*(k+1)² + (2(k+1))³
Expanding the second term, we get: 2k²*(k+1)² + 8(k+1)³
Factoring out (k+1)², we have: (k+1)²(2k² + 8(k+1))
Expanding the second term, we get: (k+1)²(2k² + 8k + 8)
Simplifying further, we have: (k+1)²(2k² + 8k + 8) = (k+1)²(2(k² + 4k + 4))
Notice that k² + 4k + 4 is the square of (k+2), so we can rewrite the equation as:
(k+1)²(2(k+2)²) = 2(k+1)²((k+2)²)
Which is equal to the RHS of the equation.
Since we have proved that the formula holds for the base case and the inductive step, we can conclude that the formula is true for all natural numbers n.