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## #1 2006-06-07 01:17:36

ant
Member
Registered: 2006-06-07
Posts: 9

### mathematical induction qu

Hey guys theres this inequalities mathematical induction question which I absolutely cannot do. Could someone please give it a go cos there arent any answers in the book. Ive tried everything I can think of, even expanding (k+1)^5  but I still cant get it, even tho I can see that its true just from common sense.

[^ = raised to the power of]

Prove:

5^n ≥ n^5        when k ≥ 5

thanks

If you're not part of the solution, you're part of the precipitate.

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## #2 2006-06-07 02:28:32

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

### Re: mathematical induction qu

check the fractual containing both leftside and rightside

do you get a hint?

X'(y-Xβ)=0

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## #3 2006-06-07 02:55:32

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,344

### Re: mathematical induction qu

The question:-

Prove that 5^n ≥ n^5 for n ≥ 5.

Proof:-
Let n=5.
LHS=5^6=15625
RHS=6^5=7776, hence LHS>RHS.

Let n=7.
LHS=5^7=78125
RHS=16807, hence, LHS>RHS.

Let 5^k >k^5 where k≥ 5.
Lets compare 5^(k+1) and (k+1)^5
5^(k+1)=5 x 5^k
(k+1)^5=k^5+5K^4+10K^3+10k^2+5K+1=5(k^5/5+k^4+2k^3+2k^2+K+1/5)
Cacelling 5 on both sides, the LHS is 5^k and
the RHS is k^5/5+k^4+2k^3+2k^2+K+1/5
5^k > k^5/5+k^4+2k^3+2k^2+K+1/5
Since the inequation is true for k+1 when it is true for k,
it is said to be true for any k≥5.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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## #4 2006-06-07 18:14:27

ant
Member
Registered: 2006-06-07
Posts: 9

### Re: mathematical induction qu

hey thanks that's really helpful - i couldnt get the step with the expansion.
at the end, could i then say

if
5^k > k^5/5 + k^4 + 2k^3 + 2k^2 + K + 1/5
then
5^k > k^5/5
and also
5^k > k^5

is that logic ok?

If you're not part of the solution, you're part of the precipitate.

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