Question

Prove the following statement by mathematical Induction.

For every integer n≥0, ᶦ⁼¹∑ⁿ⁺¹ i∙2ᶦ = n∙2ⁿ⁺²+2

Answer 1

To prove the statement by mathematical induction, we need to show that it holds true for the base case (n = 0) and then establish the inductive step to prove that if it holds true for some integer k, it also holds true for k+1.

Step-by-step explanation:

To prove the statement by mathematical induction, we need to show that it holds true for the base case (n = 0) and then establish the inductive step to prove that if it holds true for some integer k, it also holds true for k+1. Let's start with the base case:

1. When n = 0, the left-hand side (LHS) of the equation is 0∙2⁰ = 0, and the right-hand side (RHS) is 0∙2²+2 = 0, so the statement is true for n = 0.

Now, assume that the statement is true for some integer k, which means ᶦ⁼¹∑ᵏ⁺¹ i∙2ᶦ = k∙2ᵏ⁺²+2. We need to show that it holds true for k+1:

1. Using the assumption, we have ᶦ⁼¹∑ᵏ⁺¹ i∙2ᶦ = k∙2ᵏ⁺²+2. Adding (k+1)∙2ᵏ⁺¹ to both sides gives us:
2. ᶦ⁼¹∑ᵏ⁺² i∙2ᶦ+2(k+1)∙2ᵏ⁺¹ = k∙2ᵏ⁺²+2 + (k+1)∙2ᵏ⁺¹
3. Expanding the left-hand side and simplifying the right-hand side, we have:
4. 2[(k+1) + ᶦ⁼¹∑ᵏ⁺¹ i∙2ᶦ] = 2[(k+1) + k∙2ᵏ⁺²+2 + (k+1)∙2ᵏ⁺¹]
5. 2[(k+1) + ᶦ⁼¹∑ᵏ⁺¹ i∙2ᶦ] = 2[(k+1) + 2k∙2ᵏ⁺²+2 + (k+1)∙2ᵏ⁺¹]
6. 2[(k+1) + ᶦ⁼¹∑ᵏ⁺¹ i∙2ᶦ] = 2[(k+1)∙(2ᵏ⁺²+2+ 1)]
7. 2[(k+1) + ᶦ⁼¹∑ᵏ⁺¹ i∙2ᶦ] = (k+1)∙2ᵏ⁺³

Therefore, the statement holds true for k+1 as well. Since it is true for the base case and the inductive step, we can conclude that the statement holds true for every integer n ≥ 0 by mathematical induction.