Determine, with proof, for what natural numbersn the inequality
n! =<(n + 1)2^n
is satisfied.
It is true for 1=< n<=5, but Iam not able to prove it through procedure of mathematical induction.
Will anybody give me hint.
So it's false for n=6. If it's false for n, can you prove that it's false for n+1? In other words, given n! > (n+1)2^n, can you prove (n+1)! > (n+2)2^(n+1)?
As a hint, since you have:
n^2 > 3
n^2+1 > 4
n^2+2n+1 > 2n+4
(n+1)^2 > 2(n+2)
So, by induction, it's false for all n>5. And you have the result for 1 <= n <= 5, so that accounts for all the natural numbers.
Post again in this thread if you're still having trouble.
Even though you are only dealing with a few values of n,
you can complete the proof for your original inequality starting with n=5,
by proving that the inequality must be true for the next n down,
given that n is natural.
P(k)
?
P(k-1)
?
Express this as much as possible using P(k).
If then if
Hence
?
?
The graph of this quadratic is U-shaped and the roots are complex,
hence it is entirely above the x-axis
Therefore, if for some n=k, it is true for all n<k, for n natural.