document.write( "Question 986924: Please help me solve and explain how to solve this mathematical induction:\r
\n" ); document.write( "\n" ); document.write( "Use mathematical induction to prove that the following is true for every positive integer n:\r
\n" ); document.write( "\n" ); document.write( "1/4+2*(1/4)+4*(1/4)+...+[2(n-1)*(1/4)]=(4n^2-3n)/4\r
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Algebra.Com's Answer #607817 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "This problem is really botched up.  You should point it out to your\r\n" );
document.write( "teacher.  This is what you have above:\r\n" );
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\n" ); document.write( "1/4+2*(1/4)+4*(1/4)+...+[2(n-1)+1/4]=(4n²-3n)/4
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document.write( "But your 2nd and 3rd terms are not what we get when we substitute\r\n" );
document.write( "2 and 3 in the given nth term [2(n-1)+1/4]\r\n" );
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document.write( "Substituting n=1 in [2(n-1)+1/4] = [2(1-1)+1/4] = [2(0)+1/4] = 1/4,\r\n" );
document.write( "That's OK for you do have 1/4 for the 1st term.  However:\r\n" );
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document.write( "Substituting n=2 in [2(n-1)+1/4] = [2(2-1)+1/4] = [2(1)+1/4] = 9/4\r\n" );
document.write( "That's not what you have for the 2nd term, 2*(1/4), for that's 1/2\r\n" );
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document.write( "Substituting n=3 in [2(n-1)+1/4] = [2(3-1)+1/4] = [2(2)+1/4] = 17/4\r\n" );
document.write( "That's not what you have for the 3rd term, 4*(1/4), for that's 1.\r\n" );
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document.write( "Also the nth term on the left is not simplified:\r\n" );
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document.write( "So let's simplify your given nth term [2(n-1)+1/4]:\r\n" );
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document.write( "2(n-1)+1/4\r\n" );
document.write( "2n-2+1/4\r\n" );
document.write( "8n/4-8/4+1/4\r\n" );
document.write( "(8n-8+1)/4\r\n" );
document.write( "(8n-7)/4\r\n" );
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document.write( "So the left side of your sequence should be like the left side below\r\n" );
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document.write( "1/4 + 9/4 + 17/4 + ... + (8n-7)/4 = (4n²-3n)/4\r\n" );
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document.write( "The right side is OK.  But since there is a 4 in the denominator of every term\r\n" );
document.write( "on the left and also on the right side, we can just prove that 4 times the left\r\n" );
document.write( "side equals 4 times the right side.  Then after we've proved that we'll just\r\n" );
document.write( "divide both sides by 4.\r\n" );
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document.write( "So all we need do is prove this by induction:\r\n" );
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document.write( "1 + 9 + 17 + ... + (8n-7) = 4n²-3n\r\n" );
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document.write( "and then divide everything through by 4.\r\n" );
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document.write( "Induction works like this:\r\n" );
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document.write( "If a formula is assumed true for one value of n, and that assumption\r\n" );
document.write( "leads to its truth for the NEXT velaue of n, then all we have to do\r\n" );
document.write( "is prove that it is true when n=1, and then it will be true for n=2.\r\n" );
document.write( "And then the fact that it is true for n=2 will show that it is true\r\n" );
document.write( "for n=3, and so on, for all integers.\r\n" );
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document.write( "1 + 9 + 17 + ... + (8n-7) = 4n²-3n\r\n" );
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document.write( "We show that it is true when n=1\r\n" );
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document.write( "When n=1 we have just the first term 1 on the left.\r\n" );
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document.write( "On the right we have 4(1)²-3(1) = 4-3 = 1\r\n" );
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document.write( "So the formula is true when n=1\r\n" );
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document.write( "So therefore we know it is true for at least one value of n.\r\n" );
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document.write( "So we want to show that if it is assumed true for one value of n, say n=k, then\r\n" );
document.write( "it is true for the next value of n=k+1 as well.  So we assume it is true for \r\n" );
document.write( "some value of n, n=k.  So we assume this is true:\r\n" );
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document.write( "1 + 9 + 17 + ... + (8k-7) = 4k²-3k\r\n" );
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document.write( "What we need to prove now is that when we add the next term \r\n" );
document.write( "[the (k+1)st term] to both sides, like this:\r\n" );
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document.write( "1 + 9 + 17 + ... + (8k-7) + [8(k+1)-7] = 4k²-3k + [8(k+1)-7]\r\n" );
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document.write( "that we will get the same thing as if we substituted k+1 for n in the\r\n" );
document.write( "formula:  1 + 9 + 17 + ... + (8n-7) = 4n²-3n\r\n" );
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document.write( "which would be\r\n" );
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document.write( "1 + 9 + 17 + ... + [8(k+1)-7] = 4(k+1)²-3(k+1)\r\n" );
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document.write( "So essentially we want to prove that adding the (k+1)st term to the formula\r\n" );
document.write( "gives the same expression as substituting (k+1) into the formula. So we\r\n" );
document.write( "want to show that \r\n" );
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document.write( "4k²-3k + [8(k+1)-7] is equal to 4(k+1)²-3(k+1)\r\n" );
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document.write( "If we can work them both down to the same thing, then we will have proved the\r\n" );
document.write( "problem by induction:\r\n" );
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document.write( "4k² - 3k + [8(k+1)-7]  ----------  4(k+1)²-3(k+1)    \r\n" );
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document.write( "4k² - 3k + [8k+8-7]    ----------  4(k²+2k+1)-3k-3\r\n" );
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document.write( "4k²-3k+8k+1            ----------  4k²+8k+4-3k-3\r\n" );
document.write( "    \r\n" );
document.write( "4k²+5k+1               ----------  4k²+5k+1  \r\n" );
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document.write( "So they both worked out to be the same. so we have proved this:\r\n" );
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document.write( "1 + 9 + 17 + ... + (8n-7) = 4n²-3n\r\n" );
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document.write( "So all we have to do to prove your original problem,\r\n" );
document.write( "which was botched by the the 2nd and 3rd terms being\r\n" );
document.write( "wrong and the nth term not simplified, is to divide\r\n" );
document.write( "both sides of what we proved, by 4:\r\n" );
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document.write( "1/4 + 9/4 + 17/4 + ... + (8n-7)/4 = (4n²-3n)/4\r\n" );
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document.write( "I cannot imagine why you were given such a messed-up problem\r\n" );
document.write( "to prove by induction.  Please point it out to your teacher.\r\n" );
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document.write( "Edwin
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