Answer:
Proof below.
Step-by-step explanation:
Let f(n)=10+ n^2 - (n - 2)^2.
We want to prove f is even for integer n > or = 1.
Let's show f is even for the base case, n=1.
f(1)=10+1^2-(1-2)^2
f(1)=10+1-(-1)^2
f(1)=11-(1)
f(1)=10
f(1) is even.
Let's assume the for some integer k > or = 1 , we have f(k) is even. This means there is some integer m so that f(k)=2m.
Note: f(k)=10+ k^2 - (k - 2)^2.
We want to show f(k+1) is even.
f(k+1)=10+ (k+1)^2 - ([k+1] - 2)^2
f(k+1)=10+ (k+1)^2 - (k-1)^2
f(k+1)=10+ ((k+1)^2 - (k-1)^2)
That difference is a difference of squares. Let's try factoring it.
f(k+1)=10+ (k+1-[k-1])(k+1+[k-1])
f(k+1)=10+ (2)(2k)
Multiply:
f(k+1)=10+4k
Turns out we didn't even need to use our induction hypothesis since 10+4k is 2(5+2k) which makes f(k+1) even given the factor of 2 present.
This proves for all integers n>=1 that f(n) is even.
We could have this without induction then. Let's do that here below this line.
-------------------
10+ n^2 - (n - 2)^2
10+[n^2 - (n - 2)^2]
Factor difference of squares:
10+(n-(n-2))(n+(n-2))
Simplify:
10+(2)(2n-2)
Factor:
2[5+(2n-2)]
Because of the factor of 2, the given expression is always even for integer n.
