Induction - Inequalities. (1 Viewer)

[behindinfinity]

Can someone help me with this question?

Prove by mathematical induction: 2^n > 3n^2, for n ≥ 8.

 

[gurmies]

2^n > 3n

For n = 8

LHS = 256

RHS = 24

LHS > RHS, and thus inequality holds for n = 8

Assume the inequality holds for n = k, where k is any positive integer

2^k > 3k

For n = k + 1

2^(k+1)= 2(2^k)

> 2(3k)

= 6k

= 3 (k + k)

> 3 (k + 1), as k ≥ 8

Which is of the same form as for n = k

Therefore, if the inequality holds for n = k, it holds for n = k + 1.

BUT, the inequality holds for n = 8.

Therefore it holds for n = 9 and so on...

Therefore, the inequality holds for all integral n > 8

 

[Trebla]

lol, question was edited since....
2ⁿ > 3n² for n ≥ 8
For n = 8,
LHS = 2⁸
= 256
RHS = 3(8)²
= 192
LHS > RHS, so statement is true for n = 1
Assume the statement is true for n = k
2^k > 3k²
Required to prove it is true for n = k + 1
2^{(k + 1)} > 3(k + 1)²
LHS = 2^{(k + 1)}
= 2.2^k
> 2.3k² by assumption
= 6k²
> 3k² + 6k + 3
Reason:
Note that for k ≥ 8, (k - 1)² - 2 > 0 (you can see this easily by drawing a graph)
Expanding gives: k² - 2k - 1 > 0
=> k² > 2k + 1
=> 3k² > 6k + 3
=> 6k² - 3k² > 6k + 3
=> 6k² > 3k² + 6k + 3
Now back to the proof:
= 3(k² + 2k + 1)
= 3(k + 1)²
= RHS
If the statement is true for n = k, it is true for n = k +1
Since it is true for n = 8, it follows by induction that it is true for all integers n ≥ 8

 

[gurmies]

^ bastard ^

 

[Continuum]

lol, question was edited since....
2ⁿ > 3n² for n ≥ 8
For n = 8,
LHS = 2⁸
= 256
RHS = 3(8)²
= 192
LHS > RHS, so statement is true for n = 1
Assume the statement is true for n = k
2^k > 3k²
Required to prove it is true for n = k + 1
2^{(k + 1)} > 3(k + 1)²
LHS = 2^{(k + 1)}
= 2.2^k
> 2.3k² by assumption
= 6k²
> 3k² + 6k + 3
Reason:
Note that for k ≥ 8, (k - 1)² - 2 > 0 (you can see this easily by drawing a graph)
Expanding gives: k² - 2k - 1 > 0
=> k² > 2k + 1
=> 3k² > 6k + 3
=> 6k² - 3k² > 6k + 3
=> 6k² > 3k² + 6k + 3
Now back to the proof:
= 3(k² + 2k + 1)
= 3(k + 1)²
= RHS
If the statement is true for n = k, it is true for n = k +1
Since it is true for n = 8, it follows by induction that it is true for all integers n ≥ 8

Sorry to bump up the thread but I've always wondered, how come people put '=' instead of '>' in the parts I've coloured above? I always thought the latter was more correct, so I have always put '>'.

Cool emotion.

 

[Templar]

Sorry to bump up the thread but I've always wondered, how come people put '=' instead of '>' in the parts I've coloured above? I always thought the latter was more correct, so I have always put '>'.

Cool emotion.

Each use of equality or inequality refers to the equation above the one in question, not the equation to the left, hence in those circumstances equality is the valid condition.

 

[Continuum]

Ahh ok, cheers for that.