Can somebody please explain how to prove by induction 2n^2 > (n+1)^2 ,for all n>3?

Question

the question is
Prove by mathematical induction that
2n^2 > (n+1)^2
For all n > or = 3

Answer 1

Let proposition P(k) be:-
2k² > (k + 1)² for k ≥ 3

Consider P(3):-
2 x 3² = 18
(3 + 1)² = 16
Thus P(3) is true

Consider P(k + 1) which is:-
2(k + 1)² > (k + 2)² for k>3

Now 2k² > k² + 2k + 1 is true
2k² + 2k > k² + 2k + 1 + 2k is true
2k² + 2k > k² + 4k + 1 is true
2k² + 2k + 3 > k² + 4k + 4 is true
2k² + 2k + 3 > (k + 2)² is true
But 2(k + 1)² > 2k² + 2k + 3 for k > 3 thus:-
2(k + 1)² > (k + 2)² and this is proposition P(k + 1)
Thus P(3) is true and P(k + 1) is true and therefore P(n) is true for n ≥ 3

Answer 2

Okay, Im not sure this is right, but here goes!
Now, its true for 3, since 2*3^2>(3+1)^2.
Now for n+3.
First 2n^2>(n+1)^2=n^2+2n+1.
Now, subtract n^2+2n+1.
That gives you n^2-2n-1>0.
Now, insert (n+3).
That gives you (n+3)^2-2(n+3)-1>0,
Or, n^2+6n+9-2n-6-1>0,
or, n^2+4n+2>0.
Now, it shoud be clear for positive integer n,
n^2+4n+2>0.

Answer 3

Test n.

Let n = 4, then let n = 5 and try several other values for n greater than 3.

Let n = 3 and see what happens.

Then let n = 2, and let n = 1 etc. to see what happens.

Answer 4

in spite of if it incredibly is concept to be actual for n=ok, then12^ok > 7^ok + 5^ok .................(a million) Now it's going to be actual for n = ok+a million if and on condition that we are in a position to coach that 12^(ok+a million) > 7^(ok+a million) + 5^(ok+a million) right here, 12^(ok+a million) = 12*12^ok >12*(7^ok + 5^ok) making use of Eq.(a million) So 12^(ok+a million) >12*(7^ok + 5^ok) = 12*7^ok + 12*5^ok >7*7^ok + 5*5^ok when you consider that 12>7 and 12>5 = 7^(ok+a million) + 5^(ok+a million) hence, 12^(ok+a million) > 7^(ok+a million) + 5^(ok+a million)