document.write( "Question 97923This question is from textbook Intro to Real Analysis
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document.write( ": Prove that n^3 + (n+1)^3 + (n+2)^3 is divisible by 9 for all n in Natural numbers. I have tried, let mathematical induction, let P(n) be the statement, then then n^9 - n is divisible by 9. I'm stuck...\r
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Algebra.Com's Answer #71231 by aaaaaaaa(138)![]() ![]() ![]() You can put this solution on YOUR website! \"x mod y\" is short for \"remainder of the division of x by y\". Iff \"x mod y\" is 0, x is divisible by y. In our case, y = 9.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "9n^2 is always divisible by 9, so we shouldn't worry about that term. The same goes for the \"+9\" at the end, and the \"+9n\" at the middle.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Therefore, \n" ); document.write( "P(x) = \n" ); document.write( "is equivalent to \n" ); document.write( "S(x) = \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now suppose S(n) is true. I'll try to prove that S(n+1) is also true.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "But wait, if S(n) is true, then\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The inductive step is done. Now we need to check the base case of natural numbers (which is 0)\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "S(x) is therefore true for all naturals.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |