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Hence, the statement holds for the $n + 1$ case. Thus by the principle of mathematical induction $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for each $n \in \mathbb N$.
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 ...
5 Μαρ 2014 · Follow along using the transcript. Prove that 11^n - 4^n is divisible by 7 for any natural number, n. [Mathematical Induction]
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Show that the sum of the first n n positive odd integers is n^2. n2. There are several ways to solve this problem. One way is to view the sum as the sum of the first 2n 2n integers minus the sum of the first n n even integers. The sum of the first n n even integers is 2 2 times the sum of the first n n integers, so putting this all together gives.
