Question

prove that n<2^n if n>1

Answer 1

Let r and s be positive integers, then the polynomial xrs-1 is xs-1 times xs(r-1) + xs(r-2) + ... + xs + 1. So if n is composite (say r.s with 1<s<n), then 2n-1 is also composite (because it is divisible by 2s-1).

Answer 2

n! / n^n = (1/n)(2/n)(3/n)...((n-1)/n)(n/n)
= [(1 * n)/n^2] [(2 * (n-2))/n^2] * ... * [(k * (n-k))/n^2]
[I paired off the first with the last term, then the second with the second to last term, etc. Note that there are k of these terms!]

So, the proof will be complete if we can show the following claim:
If x is an integer among 1 to n, then x * (n - x) / n^2 <= 1/2.

Proof:
x * (n - x) / n^2
= x/n - x^2/n^2
= 1/4 - (1/2 - x/n)^2.
This has a maximum value of 1/4, which is less than 1/2.

Answer 3

how about using mathematical induction?

Answer 4

@patty_1CE are u familiar with mathematical induction?

Answer 5

somewhat, its a problem for my real analysis class

Answer 6

so thats it...we must use induction

Answer 7

next question... how?

Answer 8

Proof: For each positive integer \(n>1\) , let \(S(n)\) be the statement \(2^n>n\)

Basis step: \(S(2)\) is the statement \(2^2>2\). Thus \(S(1)\) is true.

Inductive step: We suppose that \(S(k)\) is true and prove that \(S(k + 1)\) is true.

Thus, we assume that\[2^k>k\]
and prove that\[2^{k+1}>k+1\]what to do now?

Answer 9

put the value

Answer 10

n>1
\(2<2^2\)

Answer 11

2<4

Answer 12

a little typo
*Thus \(S(2)\) is true.

Answer 13

note that\[2^{k+1}=2\times2^k>2k>k+1\]hence proved and the statement holds for all \(n>1\)