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30 Μαρ 2016 · I have to prove that $1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1))}{3}$ So first I did the base case which would be $1$. $1^2 = (1(2(1)-1)(2(1)+1)) / 3 1 = 3/3 1 = 1$ Which is right.
16 Απρ 2024 · Question2: Prove the following by using the principle of mathematical induction 13 + 23 + 33+ + n3 = ( ( +1)/2)^2 Let P (n) : 13 + 23 + 33 + 43 + ..+ n3 = ( ( +1)/2)^2 For n = 1, L.H.S = 13 = 1 R.H.S = (1(1 + 1)/2)^2= ((1 2)/2)^2= (1)2 = 1 Hence, L.H.S. = R.H.S P(n) is true for n = 1 Assume that P(k) is true 13 + 23 + 33 + 43 + ..+ k3 = ( ( + 1 ...
To paraphrase, the principle says that, given a list of propositions P(n), one for each n ∈ N, if P(1) is true and, moreover, P(k + 1) is true whenever P(k) is true, then all propositions are true. We will refer to this principle as mathematical induction or simply induction.
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16 Απρ 2024 · Question 5: Prove the following by using the principle of mathematical induction for all n N: 1.3 + 2.32 + 3.33 + .. n.3n = ( (2 1) 3^ ( + 1) + 3 )/4 Let P (n) : 1.3 + 2.32 + 3.33 + .. n.3n = ( (2 1) 3^ ( + 1) + 3 )/4 For n = 1, we have L.H.S =1.3 = 3 R.H.S = ( (2.1 1) 3^ (1+1) + 3)/4 = (1 3^2 + 3)/4 = (9 + 3)/4 = 12/4 = 3 Hence, L.H.S. = R.H.S...
16 Απρ 2024 · Example 1 For all n ≥ 1, prove that 12 + 22 + 32 + 42 +…+ n2 = (n(n+1)(2n+1))/6 Let P(n) : 12 + 22 + 32 + 42 + …..+ n2 = (𝑛(𝑛 + 1)(2𝑛 + 1))/6 Proving ...
