To prove 3+7+11+...+(4n − 1) = n(2n + 1) holds for an arbitrary n € Z+, in the inductive step, under our previous assumption, we prove that Check ALL that apply, i.e. you are allowed to choose ONE or MORE answers. OP(n+1): 3+7+11+. . +4(n+1)-1= 2n² + 5n +31 P(k+1): 3+7+11+...+ 4(k+ 1) − 1 = (k+1)(2(k+1)+1) n+1 P(n+1): (4(+1) − 1) = (n + 1)(2(n + 1) + 1) j=1 n+1 P(n+1): (4j-1) = (n + 1)(2(n+1) + 1) Σw-

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A: See the attachment

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A:

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A: The given problem can be solved using mathematical induction:

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A:

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A:

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A: given,

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A: Topic:- application of algebra

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A:

Q: prove its true by mathematical induction

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Q: 1. 2. Find the sum 1. Vi+v2 V2+V3 V3+v4t.+ V99+V100

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Q: Select all of the following that are equivalent to >(2) +> (3k + 1)? k=0 k=0 п+1 A. O) (3k + 3) k=0…

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Q: Prove that the statement 1 + 5 + 9 + ….. + (4n - 3) = n(2n - 1) is true.

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