document.write( "Question 1188707: Prove by induction that 2^n>2n gor every positive integer n>2.
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Algebra.Com's Answer #819856 by ikleyn(52788) $\"\"$ $\"About$

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document.write( "(1)  Base case n= 3.\r\n" );
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document.write( "     Then   = 8,  and 8 > 2*3 = 6;  so the base of induction is established.\r\n" );
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document.write( "(2)  The induction step from n to (n+1), for n > 2.\r\n" );
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document.write( "     So we assume that  > 2n  for some positive integer n > 2.\r\n" );
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document.write( "     We have   =  =  + .           (1)\r\n" );
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document.write( "     According to the induction assumption,   > 2n,  so we can continue the preceding line in this way\r\n" );
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document.write( "          =  =  +  > 2n + 2n.      (2)\r\n" );
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document.write( "     Next, we can continue this way\r\n" );
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document.write( "         2n + 2n = 2*(n+1) + 2(n-1),  and since  n > 2, the last addend is positive.\r\n" );
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document.write( "     THEREFORE,  2n + 2n > 2*(n+1)  for n > 2.         (3)\r\n" );
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document.write( "     Combining all these parts (1), (2) and (3) together, we have\r\n" );
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document.write( "           =  =  +  > 2n + 2n > 2*(n+1)  for n > 2.\r\n" );
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document.write( "     Thus the proof for the inductive step  n ---> (n+1)  is complete.\r\n" );
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document.write( "(3)  Due to the principle of Mathematical induction, the statement is proved for all positive integer n.\r\n" );
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