document.write( "Question 338564: What will be the basis step for the sum of the first n odd integers. Explain the solution. \n" ); document.write( "
Algebra.Com's Answer #242660 by Edwin McCravy(20060)\"\" \"About 
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document.write( "The nth odd positive integer is given by either\r\n" );
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document.write( "2n+1, where n begins with 0\r\n" );
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document.write( "or by\r\n" );
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document.write( "2n-1, where n begins with 1\r\n" );
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document.write( "I will use the latter, since ordinarily we begin counting with 1, not 0. \r\n" );
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document.write( "Let Sn represent the sum.\r\n" );
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document.write( "Sn = 1 + 3 + 5 + ··· + (2n-1)\r\n" );
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document.write( "The odd number just before (2n-1) is (2n-3) and the one before that is\r\n" );
document.write( "(2n-5), etc. So the above series can be written\r\n" );
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document.write( "Sn = 1 + 3 + 5 + ··· + (2n-5) + (2n-3) + (2n-1)\r\n" );
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document.write( "This is the same sum as when we reverse the terms:\r\n" );
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document.write( "Sn = (2n-1) + (2n-3) + (2n-5) + ··· + 5 + 3 + 1\r\n" );
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document.write( "Now when we add the two equations term by term, we get:\r\n" );
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document.write( "2Sn = [1+(2n-1)]+[3+(2n-3)]+[5+(2n-5)]+ ··· +[(2n-5)+5]+[(2n-3)+3]+[(2n-1)+1]\r\n" );
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document.write( "or upon simplifying\r\n" );
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document.write( "2Sn = 2n + 2n + 2n + ··· + 2n + 2n + 2n\r\n" );
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document.write( "Factoring 2 out of the right side:\r\n" );
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document.write( "2Sn = 2(n + n + n + ··· + n + n + n)\r\n" );
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document.write( "Dividing both sides by 2\r\n" );
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document.write( "Sn = n + n + n + ··· + n + n + n\r\n" );
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document.write( "There are n terms on the right, all which are n, so their sum is n×n or n².\r\n" );
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document.write( "Therefore\r\n" );
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document.write( "Sn = n²\r\n" );
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document.write( "And we see that\r\n" );
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document.write( "1 = 1 = 1²\r\n" );
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document.write( "1+3 = 4 = 2²\r\n" );
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document.write( "1+3+5 = 9 = 3²\r\n" );
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document.write( "1+3+5+7 = 16 = 4²\r\n" );
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document.write( "etc.\r\n" );
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document.write( "Edwin
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