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

Hey guys theres this inequalities mathematical induction question which I absolutely cannot do. Could someone please give it a go cos there arent any answers in the book. Ive tried everything I can think of, even expanding (k+1)^5 but I still cant get it, even tho I can see that its 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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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

check the fractual containing both leftside and rightside

do you get a hint?

**X'(y-Xβ)=0**

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,082

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.

Character is who you are when no one is looking.

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

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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