Math, asked by ks8527382407, 9 months ago

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Answered by guptapragati492

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$<front size="4"> Let P(n): 4n – 1 is divisible by 3 for each natural number n.Now, P(l): 41 – 1 = 3, which is divisible by 3 Hence, P(l) is true.Let us assume that P(n) is true for some natural number n = k.P(k): 4k – 1 is divisible by 3or 4k – 1 = 3m, m∈ N (i)Now, we have to prove that P(k + 1) is true.P(k+ 1): 4k+1 – 1= 4k-4-l= 4(3m + 1) – 1 [Using (i)]= 12 m + 3= 3(4m + 1), which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.Hence, by the principle of mathematical induction P(n) is true for all natural numbers n. $

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