document.write( "Question 650517: The sum of eight consecutive even integers; n = first integer of the eight
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Algebra.Com's Answer #407327 by swincher4391(1107)\"\" \"About 
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Let n be the first; (n+2) be the second...\r
\n" ); document.write( "\n" ); document.write( "Notice that for each kth even number, we can write it as (n+2(k-1)). This is unnecessary to solve the problem, but it may help you see where I come up with this next step:\r
\n" ); document.write( "\n" ); document.write( "So continuing on, the 8th must be (n+14)\r
\n" ); document.write( "\n" ); document.write( "So we have n+(n+2)+(n+4)+(n+6)+(n+8)+(n+10)+(n+12)+(n+14)\r
\n" ); document.write( "\n" ); document.write( "=8n + 2*(sum of 1 to 7)
\n" ); document.write( "=8n + 2*(28)
\n" ); document.write( "=8n+56
\n" ); document.write( "=8(n+7)
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\n" ); document.write( "Aside:
\n" ); document.write( "Curiously, what if it was the first 9 even numbers?\r
\n" ); document.write( "\n" ); document.write( "Then we'd have an added on (n+16).\r
\n" ); document.write( "\n" ); document.write( "This would be 9n + 2*(sum of 1 to 8)
\n" ); document.write( "= 9n + 2*(36)
\n" ); document.write( "= 9n + 72
\n" ); document.write( "= 9(n+8)\r
\n" ); document.write( "\n" ); document.write( "So, in general, the sum of k consecutive integers
\n" ); document.write( "= k(n+(k-1))
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