Extra Credit Assignment Let Pn be the statement that 2 divid
Solution
1.
a)
P(1) is the statement that:2|1^1+1
b)
P(1)=1^2+1=2
Hence, 2|P(1)
c)
Inductive hypothesis
Assume it is true for some k>=1
ie P(k) is true
d)
We need to prove ,P(k+1) is true
e)
(k+1)^2+(k+1)=k^2+1+2k+k+1=(k^2+k)+2k+2
2|k^2+k as P(k) is true
2|2k+2
Hence, P(k+1) is true
(2)
1/2+1/4=3/4=(4-1)/4
1/2+1/4+1/8=3/4+1/8=7/8=(8-1)/8
So,
1/2+1/4+...+1/2^n=(2^n-1)/2^n
b)
Let P(n) be statement that:
1/2+1/4+...+1/2^n=(2^n-1)/2^n
base case:
1/2=(2^1-1)/2
Hence, P(1) is true.
Inductive step:
Assume P(k) is true for some: k>=1
We show that: P(k+1) is true
1/2+...+1/2^n+1/2^{n+1}=(1/2+...+1/2^n)+1/2^{n+1}
=(2^n-1)/2^n+1/2^{n+1}
=(2^{n+1}-2+1)/2^{n+1}
=(2^{n+1}-1)/2^{n+1}
Hence, P(n+1) is true.
3.
One statement in Inductive step is:
\".. Because the set of the first k horses and the set of the last k horses overlap..\"
This is wrong because: k>=1 and for k=1 this does not hold true.

