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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Here is the solution to problem 2 using the Principle of Mathematical Induction. 2. Prove that _k=1^n k^2 = (n(n+1)(2n+1))/(6), n 1. Let P(n) be the statement _k=1^n k^2 = (n(n+1)(2n+1))/(6). 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) = (6)/(6) = 1 Since LHS = RHS, P(1) is true. Inductive Hypothesis: Assume P(m) is true for some positive integer m. That is, assume _k=1^m k^2 = (m(m+1)(2m+1))/(6). Inductive Step: We need to prove that P(m+1) is true, i.e., _k=1^m+1 k^2 = ((m+1)((m+1)+1)(2(m+1)+1))/(6) = ((m+1)(m+2)(2m+3))/(6). Consider the LHS of P(m+1): _k=1^m+1 k^2 = ( _k=1^m k^2 ) + (m+1)^2 By the Inductive Hypothesis, substitute _k=1^m k^2 = (m(m+1)(2m+1))/(6): (m(m+1)(2m+1))/(6) + (m+1)^2 Factor out (m+1): (m+1) [ (m(2m+1))/(6) + (m+1) ] Find a common denominator: (m+1) [ (m(2m+1))/(6) + (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: 2m^2+7m+6 = 2m^2+4m+3m+6 = 2m(m+2)+3(m+2) = (2m+3)(m+2) Substitute the factored quadratic back into the expression: (m+1) ((m+2)(2m+3))/(6) ((m+1)(m+2)(2m+3))/(6) This is the RHS of P(m+1). Thus, P(m+1) is true. By the Principle of Mathematical Induction, P(n) is true for all n 1. The proof is complete.