Use mathematical induction to show that for n >= 1, 1/(1.2) + 1/(2.3) + ... + 1/(n(n+1)) = n/(n+1) Put the steps in order to prove the equations. I. P(1) is true II. (k+1)^2 / ((k+1)(k+2)) III. 1 / (1.2) = 1 / (1+1) IV. Assume P(k) is true V. k(k+2) / ((k+1)(k+2)) + 1 / ((k+1)(k+2)) VI. 1/1.2 + 1/2.3 + ... + 1/(k(k+1)) = k/(k+1) VII. 1/1.2 + 1/2.3 + ... + 1/(k(k+1)) + 1/((k+1)(k+2)) = (k+1)/(k+2) VIII. k/(k+1) + 1/((k+1)(k+2)) IX. Basis X. (k^2+2k+1) / ((k+1)(k+2)) XI. Show P(k + 1) is true XII. (k+1) / (k+2)
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We need to use mathematical induction to show that for \(n > 1\), the following equation holds true: \[ 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n(n+1)} = \frac{n}{n+1} \] Show more…
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1. Using the principle of mathematical induction, prove that 1² + 2² + 3² + ..... + n² = (1/6){n(n + 1)(2n + 1)} for all n ∈ N. Solution: Let the given statement be P(n). Then, P(n): 1² + 2² + 3² + ..... + n² = (1/6){n(n + 1)(2n + 1)}. Putting n =1 in the given statement, we get LHS = 1² = 1 and RHS = (1/6) × 1 × 2 × (2 × 1 + 1) = 1. Therefore LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, P(k): 1² + 2² + 3² + ..... + k² = (1/6){k(k + 1)(2k + 1)}. Now, 1² + 2² + 3² + ......... + k² + (k + 1)² = (1/6) {k(k + 1)(2k + 1) + (k + 1)² = (1/6){(k + 1).(k(2k + 1)+6(k + 1))} = (1/6){(k + 1)(2k² + 7k + 6)} = (1/6){(k + 1)(k + 2)(2k + 3)} = 1/6{(k + 1)(k + 1 + 1)[2(k + 1) + 1]} ⇒ P(k + 1): 1² + 2² + 3² + ..... + k² + (k + 1)² = (1/6){(k + 1)(k + 1 + 1)[2(k + 1) + 1]} ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Adi S.
11. For any integer n ≥ 1, prove that: (a) 1 + 2 + 3 + ... + (n - 1) + n + (n - 1) + ... + 3 + 2 + 1 = n². (b) 1/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ... + 1/(n(n + 1)) = n/(n + 1). [Hint: Use the splitting identity 1/k - 1/(k + 1) = 1/k(k + 1) to rewrite the left-hand side.]
Madhur L.
(c) Prove that for any positive integer n, sum_{j=1}^{n} j(j - 1) = (n(n^2 - 1))/3 Solution Proof. By induction on n. Base case: n = 1. sum_{j=1}^{1} j(j - 1) = 1 * (1 - 1) = 0 and (n(n^2 - 1))/3 = (1(1^2 - 1))/3 = 0, so equality holds when n = 1. Inductive step: Assume that k is a positive integer and sum_{j=1}^{k} j(j - 1) = (k(k^2 - 1))/3. Show that sum_{j=1}^{k+1} j(j - 1) = ((k+1)((k+1)^2 - 1))/3. sum_{j=1}^{k+1} j(j - 1) = sum_{j=1}^{k} j(j - 1) + (k + 1)k = (k(k^2 - 1))/3 + (k + 1)k by the inductive hypothesis = (k(k^2 - 1) + 3(k + 1)k)/3 = (k(k + 1)(k - 1) + 3(k + 1)k)/3 = ((k + 1)[k(k - 1) + 3k])/3 = ((k + 1)[k^2 - k + 3k])/3 = ((k + 1)[k^2 + 2k])/3 = ((k + 1)[(k + 1)^2 - 1])/3 Therefore, sum_{j=1}^{k+1} j(j - 1) = ((k+1)((k+1)^2 - 1))/3.
Kiran P.
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