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Linear Algebra please show all work Linear Algebra please show all work Question 1 (5 points). Use mathematical induction to show that 1^2 + 2^2 + ? + n^2 = n(n + 1)(2n + 1)/6, for all n epsilon N. Solution Basis for P(1): LHS: 12 = 1 RHS: [1(1+1)(2(1)+1)]/6 = (2)(3) / 6 = 1 Basis for P(1) holds Induction: Assume P(k) holds for some integer k?1: 12 + 22 + 32 + ... + k2 + = [k(k+1)(2k+1)]/6 Goal is to establish that: 12 + 22 + 32 + ... + k2 + (k+1)2 = [(k+1)(k+2)(2(k+1)+1)]/6 = k(k+1)(2k+1) + (k+1)2 /6 = k(k+1)(2k+1) + 6(k+1)2 / 6 = (k+1)[(k(2k+1) + 6(k+1))] / 6 = (k+1)[2k2+k + 6k+6] / 6 = (k+1)[(2k2+7k+6))] / 6 = (k+1)[(2k2+4k + 3k+6))] / 6 = (k+1)[(2k(k+2) +3(k+2))] / 6 = (k+1)(k+2)(2k+3) / 6 = (k+1)(k+2)(2(k+1)+1) / 6 hence proof :).

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Linear Algebra please show all work Linear Algebra please show all work Question 1 (5 points). Use mathematical induction to show that 1^2 + 2^2 + ? + n^2 = n(n + 1)(2n + 1)/6, for all n epsilon N. Solution Basis for P(1): LHS: 12 = 1 RHS: [1(1+1)(2(1)+1)]/6 = (2)(3) / 6 = 1 Basis for P(1) holds Induction: Assume P(k) holds for some integer k?1: 12 + 22 + 32 + ... + k2 + = [k(k+1)(2k+1)]/6 Goal is to establish that: 12 + 22 + 32 + ... + k2 + (k+1)2 = [(k+1)(k+2)(2(k+1)+1)]/6 = k(k+1)(2k+1) + (k+1)2 /6 = k(k+1)(2k+1) + 6(k+1)2 / 6 = (k+1)[(k(2k+1) + 6(k+1))] / 6 = (k+1)[2k2+k + 6k+6] / 6 = (k+1)[(2k2+7k+6))] / 6 = (k+1)[(2k2+4k + 3k+6))] / 6 = (k+1)[(2k(k+2) +3(k+2))] / 6 = (k+1)(k+2)(2k+3) / 6 = (k+1)(k+2)(2(k+1)+1) / 6 hence proof :)

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Linear Algebra please show all work Linear Algebra please show all w.pdf

  • 1. Linear Algebra please show all work Linear Algebra please show all work Question 1 (5 points). Use mathematical induction to show that 1^2 + 2^2 + ? + n^2 = n(n + 1)(2n + 1)/6, for all n epsilon N. Solution Basis for P(1): LHS: 12 = 1 RHS: [1(1+1)(2(1)+1)]/6 = (2)(3) / 6 = 1 Basis for P(1) holds Induction: Assume P(k) holds for some integer k?1: 12 + 22 + 32 + ... + k2 + = [k(k+1)(2k+1)]/6 Goal is to establish that: 12 + 22 + 32 + ... + k2 + (k+1)2 = [(k+1)(k+2)(2(k+1)+1)]/6 = k(k+1)(2k+1) + (k+1)2 /6 = k(k+1)(2k+1) + 6(k+1)2 / 6 = (k+1)[(k(2k+1) + 6(k+1))] / 6 = (k+1)[2k2+k + 6k+6] / 6 = (k+1)[(2k2+7k+6))] / 6 = (k+1)[(2k2+4k + 3k+6))] / 6 = (k+1)[(2k(k+2) +3(k+2))] / 6 = (k+1)(k+2)(2k+3) / 6 = (k+1)(k+2)(2(k+1)+1) / 6 hence proof :)