(a) Use the principle of mathematical induction to show that for all integers n = 1, 1×2+2×5+3×8+cdots+n(3n-1)=n^(2)(n+1).

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(a) Use the principle of mathematical induction to show that for all integers  n >= 1, 1×2+2×5+3×8+cdots+n(3n-1)=n^(2)(n+1).

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State Proposition P(n): State the proposition P(n) that we want to prove using mathematical induction. P(n): 1×2 + 2×5 + 3×8 + ... + n(3n-1) = n^2(n+1)Verify Base Case: Verify the base case, P(1), to ensure it holds true. For n = 1, the left-hand side of the equation is 1×2 = 2. The right-hand side of the equation is 1^2(1+1) = 1×2 = 2. Since both sides are equal, P(1) is true.Assume Induction Hypothesis: Assume that P(k) is true for some positive integer k, where k >= 1. This is the induction hypothesis. Assume 1×2 + 2×5 + 3×8 + ... + k(3k-1) = k^2(k+1)Show Implication for P(k+1): Show that P(k) being true implies that P(k+1) is also true. We need to prove that 1×2 + 2×5 + 3×8 + ... + k(3k-1) + (k+1)(3(k+1)-1) = (k+1)^2(k+2)Simplify Left-hand Side: Start with the left-hand side of P(k+1) and use the induction hypothesis to simplify. 1×2 + 2×5 + 3×8 + ... + k(3k-1) + (k+1)(3(k+1)-1) = k^2(k+1) + (k+1)(3k+3-1) [Using the induction hypothesis] = k^2(k+1) + (k+1)(3k+2)Expand and Simplify: Expand and simplify the expression. = k^3 + k^2 + 3k^2 + 6k + 2k + 2 = k^3 + 4k^2 + 8k + 2Factor Expression: Factor the expression to match the right-hand side of P(k+1). = (k+1)^2(k+2)Verify Equality: Verify that the right-hand side of P(k+1) matches the simplified left-hand side. Since both sides are equal, P(k+1) is true whenever P(k) is true.Conclude by Induction: Conclude the proof by mathematical induction. Since P(1) is true and P(k) implies P(k+1), by the principle of mathematical induction, P(n) is true for all integers n >= 1.

(a) Use the principle of mathematical induction to show that for all integers $n \geq 1$ , $1\times2+2\times5+3\times8+\cdots+n(3n-1)=n^{2}(n+1).$

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(a) Use the principle of mathematical induction to show that for all integers n≥1n \geq 1n≥1, 1×2+2×5+3×8+⋯+n(3n−1)=n2(n+1).1\times2+2\times5+3\times8+\cdots+n(3n-1)=n^{2}(n+1).1×2+2×5+3×8+⋯+n(3n−1)=n2(n+1).

(a) Use the principle of mathematical induction to show that for all integers $n \geq 1$ , $1\times2+2\times5+3\times8+\cdots+n(3n-1)=n^{2}(n+1).$