a) The statement P(1) is:
1³ = (1(1 + 1)/2)²
b) To show that P(1) is true, we substitute the value of n = 1 into the equation:
1³ = (1(1 + 1)/2)²
The left-hand side of the equation is:
1³ = 1
The right-hand side of the equation is:
(1(1 + 1)/2)² = (1(2)/2)² = (2/2)² = 1² = 1
Therefore, the equation is true for n = 1.
c) The inductive hypothesis is that the statement P(k) is true for some positive integer k. In other words, assume that:
1³ + 2³ + · · · + k³ = (k(k + 1)/2)²
d) In the inductive step, we need to prove that if the statement P(k) is true, then the statement P(k + 1) is also true. In other words, we need to prove that:
1³ + 2³ + · · · + k³ + (k + 1)³ = ((k + 1)((k + 1) + 1)/2)²
e) Using the inductive hypothesis, assume that the statement P(k) is true:
1³ + 2³ + · · · + k³ = (k(k + 1)/2)²
Now, we need to prove the statement for P(k + 1):
1³ + 2³ + · · · + k³ + (k + 1)³ = ((k + 1)((k + 1) + 1)/2)²
Adding (k + 1)³ to both sides of the equation:
1³ + 2³ + · · · + k³ + (k + 1)³ = (k(k + 1)/2)² + (k + 1)³
Expanding and simplifying the right-hand side:
1³ + 2³ + · · · + k³ + (k + 1)³ = (k²(k + 1)²/4) + (k + 1)³
Combining the terms with a common denominator:
1³ + 2³ + · · · + k³ + (k + 1)³ = (k²(k + 1)² + 4(k + 1)³)/4
Simplifying further:
1³ + 2³ + · · · + k³ + (k + 1)³ = ((k + 1)(k² + 4(k + 1)))/4
Expanding and simplifying:
1³ + 2³ + · · · + k³ + (k + 1)³ = ((k + 1)(k² + 4k + 4))/4
Factoring out (k + 1)²:
1³ + 2³ + · · · + k³ + (k + 1)³ = ((k + 1)²(k + 4))/4
Simplifying further:
1³ + 2³ + · · · + k³ + (k + 1)³ = ((k + 1)((k + 1) + 1)/2)²
This is the same as the right-hand side of the statement P(k + 1), which completes the inductive step.
f) The basis step (P(1)) has been shown to be true, and the inductive step