This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Problem 2: Prove that _k=1^n k^2 = (n(n+1)(2n+1))/(6), n ≥ 1. Proof by Mathematical Induction: Step 1: Base Case (n=1). LHS: _k=1^1 k^2 = 1^2 = 1. RHS: (1(1+1)(2(1)+1))/(6) = (1 · 2 · 3)/(6) = 1. Since LHS = RHS, the formula holds for n=1. Step 2: Inductive Hypothesis. Assume the formula holds for some positive integer m: _k=1^m k^2 = (m(m+1)(2m+1))/(6) Step 3: Inductive Step. Prove the formula holds for n=m+1: _k=1^m+1 k^2 = _k=1^m k^2 + (m+1)^2 Substitute the inductive hypothesis: = (m(m+1)(2m+1))/(6) + (m+1)^2 Factor out (m+1): = (m+1) [ (m(2m+1))/(6) + (m+1) ] = (m+1) [ (2m^2+m + 6(m+1))/(6) ] = (m+1) [ (2m^2+m+6m+6)/(6) ] = (m+1) [ (2m^2+7m+6)/(6) ] Factor the quadratic 2m^2+7m+6 = (m+2)(2m+3): = ((m+1)(m+2)(2m+3))/(6) This is the RHS for n=m+1. By mathematical induction, the formula is true for all n ≥ 1. The proof is complete. Problem 3: Prove that _k=1^n (2k-1)^2 = (1)/(3)n(4n^2-1). Step 1: Expand the term (2k-1)^2. _k=1^n (2k-1)^2 = _k=1^n (4k^2 - 4k + 1) Step 2: Split the summation and apply known summation formulas. = 4 _k=1^n k^2 - 4 _k=1^n k + _k=1^n 1 Using _k=1^n k^2 = (n(n+1)(2n+1))/(6), _k=1^n k = (n(n+1))/(2), and _k=1^n 1 = n: = 4 ( (n(n+1)(2n+1))/(6) ) - 4 ( (n(n+1))/(2) ) + n = (2n(n+1)(2n+1))/(3) - 2n(n+1) + n Step 3: Factor out n and simplify. = n [ (2(n+1)(2n+1))/(3) - 2(n+1) + 1 ] = n [ (2(2n^2+3n+1))/(3) - (2n+2) + 1 ] = n [ (4n^2+6n+2)/(3) - 2n - 1 ] = n [ (4n^2+6n+2 - 3(2n+1))/(3) ] = n [ (4n^2+6n+2 - 6n - 3)/(3) ] = n [ (4n^2 - 1)/(3) ] = (1)/(3)n(4n^2-1) This matches the RHS. The proof is complete. Problem 4: Prove that n^3-n is divisible by 6 for all n ≥ 1. Step 1: Factor the expression n^3-n. n^3-n = n(n^2-1) Using the difference of squares formula, n^2-1 = (n-1)(n+1): n^3-n = n(n-1)(n+1) Step 2: Rearrange the factors. n^3-n = (n-1)n(n+1) This expression represents the product of three consecutive integers. Step 3: Demonstrate divisibility by 2 and 3. Divisibility by 2: In any set of three consecutive integers, at least one integer must be even. Therefore, their product (n-1)n(n+1) is always divisible by 2. Divisibility by 3: In any set of three consecutive integers, exactly one integer must be a multiple of 3. Therefore, their product (n-1)n(n+1) is always divisible by 3. Step 4: Conclude divisibility by 6. Since the product (n-1)n(n+1) is divisible by both 2 and 3, and 2 and 3 are coprime, the product must be divisible by 2 × 3 = 6. Thus, n^3-n is divisible by 6 for all n ≥ 1. The proof is complete.