Mathematical Induction?

Question

I understand how to do regular problems using mathematical induction, but now my teacher assigned us harder problems that we haven't learned yet.

Use mathematical induction to prove that n < 2^n whenever n is a positive integer.

Answer 1

First prove the base case, n=1: 1 < 2^1 = 2 is true.

Now, suppose that n < 2^n. We need to prove that n+1 < 2^(n+1).
We know n < 2^n, so n + 1 < 2^n + 1
< 2^n + 2^n (since for n positive 2^n > 1)
= 2^(n+1)
So n+1 < 2^(n+1) also.
So, by induction, n < 2^n for every positive integer n.

Answer 2

For n=1: n<2^n {1<2}
For n=2: n<2^n {2<4}

Assuming P(n) = n, hence P(n) < 2^n
Now, we need to prove that P(n+1) < 2^(n+1).
P(n+1) = n+1 <2^n + 1
Considering R.H.S:
2^n + 1 < 2^n.2 as 2^n > 1 (for n>0).
Hence, P(n+1) < 2^(n+1)
Proved.

Answer 3

n < 2^n
lim. n =>0
0>1
let n be e
e< 2^e
ln e < e ln2
1 < e ln2
now, let n be 1
1 < 2^1
1 < 2

It is now possible to induce that in all cases where n is an integer that n will be smaller than 2^n

Answer 4

prove n=1 is easy. (1 < 2^(1) is true)

assume n=k is true, then
Left-Hand-Side
= k+1
< 2^k + 1 ......................( k < 2^k is true)
<2^k +2^k ......................( 2^k > 1 when k > 1)
=2 (2^k)
=2^(k+1)
=Right-Hand-Side
so, k+1 < 2^(k+1)
Hence, we know that when it is true for n=k, it is true for n=k+1

By mathematical induction, it is true for n >= 1 (positive integer)

Answer 5

assume n=k is true, then
Left-Hand-Side
= k+1
< 2^k + 1 ......................( k < 2^k is true)
<2^k +2^k ......................( 2^k > 1 when k > 1)
=2 (2^k)
=2^(k+1)
=Right-Hand-Side
so, k+1 < 2^(k+1)
Hence, we know that when it is true for n=k, it is true for n=k+1

Answer 6

Its a useless chapter.

Answer 7

you must have mr. sutton ..lol