Prove that sum_k=1^n (2k-1)2 = (1)/(3)n(4n2-1), n 1.

Here are the proofs using the Principle of Mathematical Induction. 3. Prove that _k=1^n (2k-1)^2 = (1)/(3)n(4n^2-1), n 1. Step 1: Base Case (n=1) For n=1, the LHS is _k=1^1 (2k-1)^2 = (2(1)-1)^2 = 1^2 = 1. The RHS is (1)/(3)(1)(4(1)^2-1) = (1)/(3)(1)(4-1) = (1)/(3)(3) = 1. Since LHS = RHS, the statement is true for n=1. Step 2: Inductive Hypothesis Assume that the statement is true for some integer k 1. That is, assume _i=1^k (2i-1)^2 = (1)/(3)k(4k^2-1). Step 3: Inductive Step (n=k+1) We need to prove that the statement is true for n=k+1: _i=1^k+1 (2i-1)^2 = (1)/(3)(k+1)(4(k+1)^2-1) Consider the LHS: _i=1^k+1 (2i-1)^2 = (_i=1^k (2i-1)^2) + (2(k+1)-1)^2 Using the inductive hypothesis, we substitute _i=1^k (2i-1)^2 = (1)/(3)k(4k^2-1): = (1)/(3)k(4k^2-1) + (2k+2-1)^2 = (1)/(3)k(4k^2-1) + (2k+1)^2 Factor 4k^2-1 as (2k-1)(2k+1): = (1)/(3)k(2k-1)(2k+1) + (2k+1)^2 Factor out (2k+1): = (2k+1)[(k(2k-1))/(3) + (2k+1)] = (2k+1)[(2k^2-k + 3(2k+1))/(3)] = (2k+1)[(2k^2-k+6k+3)/(3)] = (2k+1)[(2k^2+5k+3)/(3)] Factor the quadratic 2k^2+5k+3: 2k^2+5k+3 = 2k^2+2k+3k+3 = 2k(k+1)+3(k+1) = (2k+3)(k+1) Substitute this back into the expression: = (2k+1)((k+1)(2k+3))/(3) = (1)/(3)(k+1)(2k+1)(2k+3) Now, let's expand the RHS we want to achieve: (1)/(3)(k+1)(4(k+1)^2-1) = (1)/(3)(k+1)(4(k^2+2k+1)-1) = (1)/(3)(k+1)(4k^2+8k+4-1) = (1)/(3)(k+1)(4k^2+8k+3) Factor the quadratic 4k^2+8k+3: 4k^2+8k+3 = 4k^2+2k+6k+3 = 2k(2k+1)+3(2k+1) = (2k+1)(2k+3) So, the RHS is: = (1)/(3)(k+1)(2k+1)(2k+3) Since LHS = RHS, the statement is true for n=k+1. By the Principle of Mathematical Induction, the statement _k=1^n (2k-1)^2 = (1)/(3)n(4n^2-1) is true for all integers n 1. 4. Prove that n^3 - n is divisible by 6 for all n 1. Step 1: Base Case (n=1) For n=1, 1^3 - 1 = 1 - 1 = 0. Since 0 is divisible by 6 (0 = 6 × 0), the statement is true for n=1. Step 2: Inductive Hypothesis Assume that the statement is true for some integer k 1. That is, assume k^3 - k is divisible by 6. This means k^3 - k = 6m for some integer m. Step 3: Inductive Step (n=k+1) We need to prove that (k+1)^3 - (k+1) is divisible by 6. Consider the expression: (k+1)^3 - (k+1) Expand (k+1)^3: = (k^3 + 3k^2 + 3k + 1) - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1 Rearrange the terms to use the inductive hypothesis: = (k^3 - k) + 3k^2 + 3k = (k^3 - k) + 3k(k+1) By the inductive hypothesis, k^3 - k is divisible by 6. So, we can write k^3 - k = 6m for some integer m. Now consider the term 3k(k+1). The product of two consecutive integers, k(k+1), is always an even number. This is because either k is even or k+1 is even. Therefore, k(k+1) = 2p for some integer p. Substitute this into 3k(k+1): 3k(k+1) = 3(2p) = 6p So, 3k(k+1) is also divisible by 6. Now substitute these back into the expression for (k+1)^3 - (k+1): (k+1)^3 - (k+1) = (k^3 - k) + 3k(k+1) = 6m + 6p = 6(m+p) Since m and p are integers, their sum m+p is also an integer. Thus, (k+1)^3 - (k+1) is a multiple of 6, which means it is divisible by 6. By the Principle of Mathematical Induction, the statement n^3 - n is divisible by 6 is true for all integers n 1.