Find and prove an inequality relating 100n and n3 Since 100n

Find and prove an inequality relating 100n and n³.

Since

100n

and

n³

for

n = 1, 2, 3, . . . 9, 10, 11

are

110, 200, 300, . . . 900, 1000, 1100

and

1, 8, 37, . . . 729, 1000, 1331

respectively, our conjecture is that

100n n³,

for all

n   .

Let

P(n)

denote the statement that

100n n³.

P(10)

is the statement that 100

, which is true.

Assume that

P(k)

is true. Thus, our induction hypothesis is

100k  ? = k³,

for some

k   .

We want to use this to show that

P(k + 1)

is true. Now,

100(k + 1) = 100k +   k + k² k³ + 3k² + 3k + 1 =

.

Thus,

P(k + 1)

follows from

P(k),

and this completes the induction step. Having proven the above steps, we conclude by the Principle of Mathematical Induction that

P(n)

is true.

Solution

Hi :)

We are doing a proof by induction.
we have to prove that 100n<=n^3 for all n>=10
Two steps.

First, we prove the statement is true for n=10:
100 * 10 10^3
true, because
1000 = 1000

Second step. Prove that if you accept the statement to be true for some k, then it must still be true for k+1.
100k k^3 is true for some k>=10, so if this statement is taken as being true; can we use it to prove:
100(k+1) (k+1)^3 ?

expanding the inequality:

100k + 100 k^3 + 3k^2 + 3k + 1

since we know that 100k k^3, all we have left to do is prove that
100 3k^2 + 3k + 1
since we only deal with k 10, we have
3k^2 300 (plugging in k=10)
3k 30
therefore (3k2 + 3k + 1) 331
Since 100 is always smaller than 331, the statement
100k + 100 k^3 + 3k^2 + 3k + 1
it is also be true, whenever 100k k^3

l hope it helps :)