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Comp4200 0.1f: The set of all integers that are equal to one added to that number. 0.1e: The set of all strings comprising of 0’s and 1’s and every string is a palindrome. 0.6d: Range R={6,7,8,9,10} Domain D=X*Y={(1,6),(1,7),(1,8),(1,9),(1,10),(2,6),(2,7),(2,8),(2,9),(2,10),(3,6),(3,7),(3,8),(3,9),(3,10 ),(4,6),(4,7),(4,8),(4,9),(4,10),(5,6),()5,7,(5,8),(5,9),(5,10),} 0.6e: f(4)=7 g(4,f(4))=g(4,7)=8 1： (1) base case: when n=1. Left = 1/2, right=1/2. It is true. (2) Assume when n = k, left side equal right side (3) Prove: when n=k+1 left still equal to right side When n=k+1, left side= (k/k+1)+(1/(k+1)(k+2)) =[k(k+1)+1]/(k+1)(k+2) =(k+1)/(k+2)= right So for all n belong to Z+ 2: (1) base case: when n= 1, left side is 1,meanwhile, right side is 1 so the base case is satisfied. (2) Assume n= k, this formula is still correct (3) Prove: when n = k+1 left= 1^3+2^3+...+k^3+(k+1)^3 =[k^2(k+1)^2]/4+(k+1)^3 =(k+1)^2[(k^2)/4+(k+1)] ={[(k+1)(k+2)]^2}/4 ={[(k+1)(k+1+1)]/2}^2 =right So for all n belong Z+