Prove that \[7n+3\le2^{n}\] for all \[n \ge6\]
Proof by induction!
Let's see if the relation is valid for \(n=6\), \[7(6)+3\le 2^6 \implies 45\le 64\] TRUE! Now, assume it's true for n=k, we have to prove it's also true for n=k+1.
Prove for n = 6: 42 + 3 <= 2^6 45<=64 True! Suppose it's true for n: 7n + 3 <= 2^n prove for n+1 7(n+1) + 3 <= 2^(n+1) 7n + 3 + 7 <= 2 * 2^n 7n + 3 + 7 <= 2^n + 2^n (7n + 3) + 7 <= 2^n + 2^n Now we said suppose it's true for n: therefore 7n + 3<= 2^n Now we must prove 7<=2^n Since we know it's true for n>=6: 2^6 > 7 Therefore 7<=2^n is true for n>=6 that proves it for n+1 Therefore it's always true
well, not always. true for n>=6
2 * 2^n = 2^n + 2^n ?
You didnt factor anything there right
its just a basic arithmetic fact
like 3*2^n would be 2^n + 2^n + 2^n
Damn it! I got confused! I need a paper :P
You can factor it, yes. 2*2^n = (1 + 1) * 2^n = 2^n + 2^n
ah ok
Mr Math, check if my proof makes sense please? I've never really done an inequality induction before
Yeah, slaaibak's work is correct.
I dont see how its proof
http://en.wikipedia.org/wiki/Mathematical_induction basically if you prove for the base case, then thereafter if you prove for n+1, you prove for all natural numbers. since you first proof for 6, then 6+1=7, then 7+1 and it goes on forever
Right
I just dont get it lol
so wait
We assumed \(7k+3\le 2^k\), we want to prove it for n=k+1, that's \[(7k+3)+7\le 2^k+2^k \text{, but we have } 7k+3\le 2^k \implies 7\le 2^k\]. \[\text{ This is true for }k\ge3.\]
we assume 7n + 3 <= 2^n is true
And hence the relation is true for all integers \(k\ge 6\).
@mwmnj, You probably need to look up proof by induction and "its proof". :D
Is that it?
Cause there isnt an intuitive obvious proof im seeing when i do these
Its just a pattern im following
Mathematical induction is one of the greatest methods of proofs. http://en.wikipedia.org/wiki/Mathematical_induction
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