x > 0; n integer
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Sounds like induction to me :P
Let n=1.
Then x-1 + x-2 = 2x-3. x>0, so…
um...
Your statement is false.
Assume it is true.
Then it should be true for n=1, x=3.
But
3^1 > (3-1)^1 + (3-2)^1
3 > 2 + 1
which is obviously false.
So the assumption must be false. But the assumption is the statement, so the statement is false.
What did you mean to say?
Let n=1.
Then x-1 + x-2 = 2x-3. x>0, so…
um...
Your statement is false.
Assume it is true.
Then it should be true for n=1, x=3.
But
3^1 > (3-1)^1 + (3-2)^1
3 > 2 + 1
which is obviously false.
So the assumption must be false. But the assumption is the statement, so the statement is false.
What did you mean to say?
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x^n > (x +1)^N + (x+2)^n
Prove true for a number
(2)^1> (2-1)^1 + (2-2)^1
2>1
therefore statement is true for 1
Assume true for x^k therefore
x^k > (x-1)^k + (x-2)^k
Prove true for k+1
x^k+1 > (x-1)^k+1 +(x-2)^k+1
LHS (X-1)k+1 + (x-2)^k+1
multiply brackets and get quadratic and solutions you would see that the statement is true
Prove true for a number
(2)^1> (2-1)^1 + (2-2)^1
2>1
therefore statement is true for 1
Assume true for x^k therefore
x^k > (x-1)^k + (x-2)^k
Prove true for k+1
x^k+1 > (x-1)^k+1 +(x-2)^k+1
LHS (X-1)k+1 + (x-2)^k+1
multiply brackets and get quadratic and solutions you would see that the statement is true