document.write( "Question 922472: I can't find an equation for the following ordered pairs (1,1), (2,5), (3,14), (4,30) etc. I know that I am adding the square of each x-value number to the previous y-value but I'm confused about the equation. \n" ); document.write( "

Algebra.Com's Answer #559594 by Edwin McCravy(20055) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "As you observed, the nth term is the sum of the first n squares. \r\n" );
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document.write( "1 = 1² \r\n" );
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document.write( "5 = 1²+2² \r\n" );
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document.write( "14 = 1²+2²+3² \r\n" );
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document.write( "30 = 1²+2²+3²+4²\r\n" );
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document.write( "So the nth term is 1²+2²+3²+...+n²\r\n" );
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document.write( "That should be related to the sum 1+2+3+...+n\r\n" );
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document.write( "1+2+3+...+n is the sum of an arithmetic series with a₁=1 and d=1\r\n" );
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document.write( "Using the formula for the sum of the an arithmetic series to n terms:\r\n" );
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document.write( "1+2+3+...+n =  or \r\n" );
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document.write( "Let's divide each term of our sequence by the sum of the \r\n" );
document.write( "first n natural numbers, and see if we get a recognizable\r\n" );
document.write( "pattern:\r\n" );
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document.write( "1/(1) = 1\r\n" );
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document.write( "5/(1+2) = 5/3\r\n" );
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document.write( "14/(1+2+3) = 14/6 = 7/3\r\n" );
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document.write( "30/(1+2+3+4) = 30/10 = 3\r\n" );
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document.write( "Now if we write 1 as 3/3 and 3 and 9/3 we do have a \r\n" );
document.write( "recognizable pattern\r\n" );
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document.write( "3/3, 5/3, 7/3, 9/3, ...\r\n" );
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document.write( "The numerators go 3,5,7,9,... each of which is 1 more than 2,4,6,8,...\r\n" );
document.write( "which has nth term 2n\r\n" );
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document.write( "So 3,5,7,9, has nth term 2n+1\r\n" );
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document.write( "So 3/3, 5/3, 7/3, 9/3 has nth term (2n+1)/3\r\n" );
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document.write( "Since we got that sequence by DIVIDING our sequence by the sums of the \r\n" );
document.write( "first n natural numbers, the nth term of our sequence is gotten by \r\n" );
document.write( "MULTIPLYING those two nth terms.  So the nth ordered pair is given\r\n" );
document.write( "by the equation:\r\n" );
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document.write( "or\r\n" );
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document.write( "Edwin

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