# hw1 ```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+
```