Please hlep **Prove the following statement by mathematical induction.**For every integer \( n \geq 1 \),\[\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}\]**Proof (by mathematical induction):** Let \( P(n) \) be the equation\[\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]We will show that \( P(n) \) is true for every integer \( n \geq 1 \).**Show that \( P(1) \) is true:** Select \( P(1) \) from the choices below.- \( P(1) = \frac{1}{1 \cdot 2} \)- \( \frac{1}{1 \cdot 2} = \frac{1}{1 + 1} \)- \(\frac{1}{1 \cdot 2} = \frac{1}{2}\)The selected statement is true because both sides of the equation equal the same quantity.**Show that for each integer \( k \geq 1 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:**Let \( k \) be any integer with \( k \geq 1 \), and suppose that \( P(k) \) is true. Select the expression for the left-hand side of \( P(k) \) from the choices below.- \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k + 1)} \)- \(\frac{k}{k + 1}\)- \(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k ### Mathematical Induction Proof**Inductive Hypothesis:** The hypothesis states that the two sides of P(k) are equal.**Goal:** Show that P(k + 1) is true. The equation for P(k + 1) is:\[\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k \cdot (k + 1)} + \frac{1}{(k + 1)(k + 2)} = \frac{k + 1}{(k + 1) + 1}\]**Task:** Determine which choice shows the result of applying the inductive hypothesis to the expression on the left-hand side of P(k + 1).**Options:**1. \(\frac{k}{k + 1} + \frac{1}{(k + 1)(k + 2)}\)2. \(\frac{1}{k + 1} + \frac{k + 1}{1(k + 2)}\)3. \(\frac{1}{k + 1} + \frac{1}{(k + 1)(k + 2)}\)4. \(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k \cdot (k + 1)} + \frac{1}{(k + 1)(k + 2)}\)The selected choice is:- \(\frac{k}{k + 1} + \frac{1}{(k + 1)(k + 2)}\)**Conclusion:**When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal:\[\frac{k + 1}{(k + 1) + 1}\]Thus, P(k + 1) is true, which completes the inductive step.**Summary:**Both the basis and the inductive steps have been proved, completing the proof by mathematical induction.

Answered: Image /qna-images/answer/93f1652c-d728-4362-8d3e-94dc5d574eb3.jpg

Prove the following statement by mathematical induction. 1 1 1 1 For every integer n2 1, 1-2 2-3 3-4 n(n + 1) n +1 Proof (by mathematical induction): Let P(n) be the equation 1 1 1 1 1-2 2-3 3- 4 n(n + 1) n+1 We will show that P(n) is true for every integer n2 1. Show that P(1) is true: Select P(1) from the choices below. 1 O P(1) 1-2 1 O P(1) 1 +1 1 1-2 1 1 1 1 1-2 2-3 3. 4 1-2 1 +1 1 1 1- 2 1(1 + 1) 1+1 The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer k 2 1, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 1, and suppose that P(k) is true. Select the expression for the left-hand side of P(k) from the choic 1 1 1- 2 k(k + 1) 1 1 1.2 2-3 3. 4 1 1 1 1-2 2-3 3. 4 k(k + 1) 1 1 1 2 -3 k(k + 1) 1-2 3.4 1 The right-hand side of P(k) is *(k +1) [The inductive hypothesis states that the two sides of P(k) are egual.]