Question 1188707
.
Prove by induction that 2^n > 2n {{{cross(gor)}}} <U>for</U> every positive integer n > 2.
~~~~~~~~~~~~~~~~~


<pre>
(1)  Base case n= 3.

     Then  {{{2^3}}} = 8,  and 8 > 2*3 = 6;  so the base of induction is established.



(2)  The induction step from n to (n+1), for n > 2.


     So we assume that {{{2^n}}} > 2n  for some positive integer n > 2.


     We have  {{{2^(n+1)}}} = {{{2*2^n}}} = {{{2^n}}} + {{{2^n}}}.           (1)


     According to the induction assumption,  {{{2^n}}} > 2n,  so we can continue the preceding line in this way

         {{{2^(n+1)}}} = {{{2*2^n}}} = {{{2^n}}} + {{{2^n}}} > 2n + 2n.      (2)


     Next, we can continue this way

         2n + 2n = 2*(n+1) + 2(n-1),  and since  n > 2, the last addend is positive.


     THEREFORE,  2n + 2n > 2*(n+1)  for n > 2.         (3)


     Combining all these parts (1), (2) and (3) together, we have

          {{{2^(n+1)}}} = {{{2*2^n}}} = {{{2^n}}} + {{{2^n}}} > 2n + 2n > 2*(n+1)  for n > 2.


     Thus the proof for the inductive step  n ---> (n+1)  is complete.



(3)  Due to the principle of Mathematical induction, the statement is proved for all positive integer n.
</pre>

Solved.