Please follow exactly the solution steps of the 2nd image to answer my question. The way you answer my question should follow the steps on that image Learning Target R4 Core: Given a statement to be proven by (weak) induction, I can state and prove the base case,state the inductive hypothesis, and outline the proof. I can describe the subtle difference between 'weak' and 'strong'induction.For each statement below, complete each of the following parts:. State and prove the base caseState the inductive hypothesis• Outline how the rest of the proof would go, that is, tell me roughly what will happen to complete theproof, identifying how this section of the proof will begin and ultimately what it will look like whenyou're done (as modeled in class videos); you do not need to actually do the complete proof.You might not be able to prove these claims with what we learned in this class, and that's okay. You're justsetting it up!Remember if you have a question about this, you can email/send me a message on Teams. I want tohelp you succeed!1. Show that 2! 4! 6! (2n)! ≥ ((n + 1)!)" for n € Zt.2. Show that a + (a+d) + (a+2d) + (a +3d) ++ (a + (n − 1)d) =n(2a + (n-1)d)2for n E Zt. 3:47Let's prove the two statements using mathematical induction:(1) Proof by Mathematical Induction:We want to prove that for all positive integers n, the following identity holds:(n(n+1))².1³ +2³+...+n³ =Base Case (n= 1):When n = 1, the left-hand side (LHS) is 1³ = 1, and the right-hand side(RHS) is (1(1+1))² = 1² = 1. The identity holds for n = 1.Inductive Hypothesis:Assume that the identity holds for some positive integer k, i.e.,Inductive Step:We need to prove that the identity also holds for k + 1, i.e.,1³ +2³+...+k³ + (k+ 1)³= = ( (k + 1)(k+1+1)) ².1³ +2³+...+k³ =We can start from the assumption and add (k+1)³ to both sides:= (k (k + ¹)² + (Now, we can simplify the RHS:(k (* + ¹)) ² +1³ +2³+...+³+ (k+ 1)³= |k²(k+ 1)²4Step2b)= (k( k + ¹)) ².k² (k+ 1)²4Now, let's manipulate the LHS:+(k+ 1)³=To prove the identity, we need to show that the LHS and RHS are equal:= ( (* + 1)(k + 2)) ².+ (k+ 1)³=+ (k+ 1)³=k²(k+ 1)²+k² (k+ 1)²4+(k+1)³.+ (k+1)³.154(k+1)³ k²(k+ 1)² + 4(k+1)³We can factor out (k+ 1)² from the numerator:(k+ 1)²(k² + 4(k+ 1)) (k+ 1)² (k² + 4k+ 4) (k+ 1)²(k+ 2)²444@This is the same as the RHS, so the identity holds for k +1, completing theinduction step.By mathematical induction, we have shown that the identity 13 +23+...+n³=3 = ((n+1)) ² ₁ holds for all positive integers n.(2) Proof that any postage greater than or equal to 23 can be made usingonly 5-cent and 7-cent stamps:√xDO8

Question

Please follow exactly the solution steps of the 2nd image to answer my question. The way you answer my question should follow the steps on that image Learning Target R4 Core: Given a statement to be proven by (weak) induction, I can state and prove the base case,state the inductive hypothesis, and outline the proof. I can describe the subtle difference between 'weak' and 'strong'induction.For each statement below, complete each of the following parts:. State and prove the base caseState the inductive hypothesis• Outline how the rest of the proof would go, that is, tell me roughly what will happen to complete theproof, identifying how this section of the proof will begin and ultimately what it will look like whenyou're done (as modeled in class videos); you do not need to actually do the complete proof.You might not be able to prove these claims with what we learned in this class, and that's okay. You're justsetting it up!Remember if you have a question about this, you can email/send me a message on Teams. I want tohelp you succeed!1. Show that 2! 4! 6! (2n)! ≥ ((n + 1)!)&#34; for n € Zt.2. Show that a + (a+d) + (a+2d) + (a +3d) ++ (a + (n − 1)d) =n(2a + (n-1)d)2for n E Zt. 3:47Let's prove the two statements using mathematical induction:(1) Proof by Mathematical Induction:We want to prove that for all positive integers n, the following identity holds:(n(n+1))².1³ +2³+...+n³ =Base Case (n= 1):When n = 1, the left-hand side (LHS) is 1³ = 1, and the right-hand side(RHS) is (1(1+1))² = 1² = 1. The identity holds for n = 1.Inductive Hypothesis:Assume that the identity holds for some positive integer k, i.e.,Inductive Step:We need to prove that the identity also holds for k + 1, i.e.,1³ +2³+...+k³ + (k+ 1)³= = ( (k + 1)(k+1+1)) ².1³ +2³+...+k³ =We can start from the assumption and add (k+1)³ to both sides:= (k (k + ¹)² + (Now, we can simplify the RHS:(k (* + ¹)) ² +1³ +2³+...+³+ (k+ 1)³= |k²(k+ 1)²4Step2b)= (k( k + ¹)) ².k² (k+ 1)²4Now, let's manipulate the LHS:+(k+ 1)³=To prove the identity, we need to show that the LHS and RHS are equal:= ( (* + 1)(k + 2)) ².+ (k+ 1)³=+ (k+ 1)³=k²(k+ 1)²+k² (k+ 1)²4+(k+1)³.+ (k+1)³.154(k+1)³ k²(k+ 1)² + 4(k+1)³We can factor out (k+ 1)² from the numerator:(k+ 1)²(k² + 4(k+ 1)) (k+ 1)² (k² + 4k+ 4) (k+ 1)²(k+ 2)²444@This is the same as the RHS, so the identity holds for k +1, completing theinduction step.By mathematical induction, we have shown that the identity 13 +23+...+n³=3 = ((n+1)) ² ₁ holds for all positive integers n.(2) Proof that any postage greater than or equal to 23 can be made usingonly 5-cent and 7-cent stamps:√xDO8<

Accepted Answer

To prove the inequality  for n∈Z+, ﻿we can use mathematical induction.Base Case: Let's check the…

1. Show that 2! 4! 6! (2n)! ≥ ((n + 1)!)" for ne z+. . ...