How can I prove by induction that:
$\displaystyle n^2 > 3n$ for all n>=4
I do:
n=k
P(k) = k^2 > 3k
P(k+1) = (k+1)^2 > 3(k+1)
How to solve
P(k) --> P(k+1) to prove that P(n) is truth ???
What you did was take an any expression to prove that (k+1)^2 > 3k + 3 ?
For example
Prove that $\displaystyle 2^{2n} -1$ is divisible by 3
I do:
$\displaystyle 2^{2k} - 1 = 3m$
$\displaystyle 2^{2k} = 3m + 1$
$\displaystyle 2^{2k+2} = 12m+4$
$\displaystyle 2^{2k+2} -1 = 3(4m+1)$
4m + 1 is divisible by 3 then $\displaystyle 2^{2n} - 1$ also divisible by 3
Correct ?
I don't understand your question. Can you rephrase?
That is correct, but you might want to say where you're using the induction hypothesis.For example
Prove that $\displaystyle 2^{2n} -1$ is divisible by 3
I do:
$\displaystyle 2^{2k} - 1 = 3m$
$\displaystyle 2^{2k} = 3m + 1$
$\displaystyle 2^{2k+2} = 12m+4$
$\displaystyle 2^{2k+2} -1 = 3(4m+1)$
4m + 1 is divisible by 3 then $\displaystyle 2^{2n} - 1$ also divisible by 3
Correct ?