Prove by mathematical induction that 13+23+33 ... + n3=[n(n+1)/2]2.
Get the complete, step-by-step math solution for: "Prove by mathematical induction that 1³ + 2³ + 3³ ... + n³ = [n(n+1)/2]².". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Base Case (n=1)
We begin by verifying the base case for n=1 .Wesubstituten=1 into both sides of the given equation and check if they are equal. The left side becomes 13=1. The right side becomes [21(1+1)]2=[21(2)]2=[1]2=1. Since both sides are equal, the base case holds true.
Step 2: Inductive Hypothesis
Next, we assume that the statement P(k) is true for some arbitrary positive integer k This means we assume that the sum of the first k cubes is equal to [2k(k+1)]2. This assumption is called the inductive hypothesis.
Step 3: Inductive Step (P(k+1))
Now, we need to prove that the statement P(k+1) is true, assuming P(k) is true. This means we need to show that the sum of the first k+1 cubes is equal to [2(k+1)(k+2)]2.
Step 4: Substitute Inductive Hypothesis
We start with the left-hand side of the P(k+1) statement. Using our inductive hypothesis, we replace the sum 13+23+⋯+k3 with its assumed equivalent, [2k(k+1)]2 Then we add the (k+1)th term, which is (k+1)3.
Step 5: Factor and Simplify
We factor out (k+1)^2 from the expression. Then, we find a common denominator for the terms inside the square brackets and combine them. The numerator k2+4k+4 is a perfect square trinomial, which can be factored as (k+2)^2$. Finally, we rewrite the entire expression as a single squared term.