Answer
(a) False.
(b) True.
(c) True.
(d) False.
(e) True.
(f) True
Work Step by Step
(a) False. $p(11)=121=11\times11$
(b) True. When $n=2$, $2^2\gt2$, assume $k^2\gt k$ is true, we have $(k+1)^2=k^2+2k+1\gt k+2k+1\gt k+1$ which is also true, thus we proved the statement through mathematical induction.
(c) True. When $n=1$, $2^{3}+1=9$ is divisible by 3. Assume when $n=k$, $2^{2k+1}+1=3m$ is divisible by 3 with $m$ as an integer. When $n=k+1$, we have $2^{2k+3}+1=4(3m-1)+1)=12m-3=3(4m-1)$ is also divisible by 3. Thus we proved the statement through mathematical induction.
(d) False. When $n=2$, $2^3=8$, $(2+1)^2=9$, and $8\lt9$
(e) True. When $n=2$, $2^3-2=6$ is divisible by 3. Assume when $n=k$, $k^2-k=3m$ is divisible by 3 (m is an integer), we have when $n=k+1$, $(k+1)^3-(k+1)=k^3+3k^2+3k+1-k-1=3m+3k^2+3k$ is also divisible by 3.
Thus we proved the statement through mathematical induction.
(f) True!
Let P(n): "$n^3-6n^2+11n = 3k$" for all $n>= 1$.
Step 1 (Base case): P(1)=$1^3-6(1)^2+11(1) = 6$
So, P(1) is true!
Step 2(Inductive step):
Assume that P(n) is true for all n>=1, then we want to show that P(n+1) is also true.
$(n+1)^3-6(n+1)^2+11(n+1)=n^3+3n^2+3n+1-6(n^2+2n+1)+11n+1$
=$(n^3-6n^2+11n) +3n^2+3n-12n-6+12$
=$3k+3n^2+3n-12n+6$
=$3(k+n^2-9n+2)$
So P(n+1) is true (when P(n) is true) for all n>=1
Conclusion: P(n) is divisible by 6 for all n>=1.
