Posted by **Dina** on Friday, September 21, 2007 at 2:00pm.

If 2^n>n^2 and n>5, then 2^n+1>(n+1)^2

Proof: Assuming that 2^n>n^2 then I can say that 2*2^2>2*n^2 = 2^n+1>2n^2

If I can show that 2n^2>(n+1)^2 then I will be done by transitivity. So 2n^2>(n+1)^2? then 2n^2>n^2+2n+1? then n^2>2n+1, hence n^2-2n-1>0, and if I add 2 then n^2-2n+1>2... after this I do not know what to do. If you can see that I did something wrong somewhere along the proof please let me know.

- Abstract Algebra -
**bobpursley**, Friday, September 21, 2007 at 2:34pm
Proof: Assuming that 2^n>n^2 then I can say that 2*2^n>2*n^2 = 2^n+1>2n^2

**yes, correct**

If I can show that 2n^2>(n+1)^2 then I will be done by transitivity.

So 2n^2>(n+1)^2?

then 2n^2>n^2+2n+1? **
**

then n^2>2n+1, now, n>5, so n= 5+m, where m is any positive integer.

n^2 >2n+1

(5+m)^2>2(5+m)+1

25+2m + m^2>10+2m+1

m^2>-14, and since m is any postive integer, it is proved.

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