document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1080223>Question 1080223</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need help with this mathematical induction to show that the given statement is true for all natural numbers,\r<BR>\n" );
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document.write( "3 + 5 + 7 +...+ (2n+1) = n(n+2)\r<BR>\n" );
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document.write( "So far I have the following:\r<BR>\n" );
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document.write( "Prove Basis n=1<BR>\n" );
document.write( "n(n+2)<BR>\n" );
document.write( "= 1(1+2)<BR>\n" );
document.write( "= 3\r<BR>\n" );
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document.write( "Therefore the statement is true for n=1.\r<BR>\n" );
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document.write( "When n=k assume<BR>\n" );
document.write( "3 + 5 + 7 +...+ (2n+1) = n(n+2)\r<BR>\n" );
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document.write( "replace n with k<BR>\n" );
document.write( "3 + 5 + 7 +...+ (2k+1) = k(k+2)\r<BR>\n" );
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document.write( "Then n = k+1 must be proven, meaning replace k terms with (k+1)\r<BR>\n" );
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document.write( "This is where I get lost, I don't know where/how I should replace with (k+1) despite how simple it sounds\r<BR>\n" );
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document.write( "I get <BR>\n" );
document.write( "3 + 5 + 7 +...+ (2(k+1)+1) = (k+1)((k+1)+2)\r<BR>\n" );
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document.write( "But some sites show a funky set up where the right/first part of the equation gets another (k+1) replacement but don't explain how. I'm sure the left hand side (k+1)((k+1)+2) is correct. I know that after that step you must replace the 3 + 5 + 7 ... etc. but I can't get that far since I'm stuck on the (k+1) replacement! Any help/advice is appreciated and I don't need it solved just some guidance! Thank you! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #694420 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=694420\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You have the right idea. You need to connect the (n+k)th case to the kth case somehow. Here's one way to do just that. I'll start where you left off.\r<BR>\n" );
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document.write( "Before I start, I'll refer back to this identity<BR>\n" );
document.write( "3 + 5 + 7 +...+ (2k+1) = k(k+2) <BR>\n" );
document.write( "which will be used later. \r<BR>\n" );
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document.write( "Onto the problem. Let's simplify things a bit<BR>\n" );
document.write( "3 + 5 + 7 +...+ (2(k+1)+1) = (k+1)((k+1)+2) <BR>\n" );
document.write( "3 + 5 + 7 +...+ (2k+3) = (k+1)(k+3)<BR>\n" );
document.write( "3 + 5 + 7 +...+ (2k+3) = k^2+4k+3<BR>\n" );
document.write( "3 + 5 + 7 +...+ (2k+1) + (2k+3) = k^2+4k+3\r<BR>\n" );
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document.write( "Notice how the portion \"3 + 5 + 7 +...+ (2k+1)\" is buried in the left side of the last equation above. This works because the (k+1)th case is really just the result of taking the kth case and adding on the next term. \r<BR>\n" );
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document.write( "We can replace all of that with the right side of the identity mentioned earlier.\r<BR>\n" );
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document.write( "Take note of the color coding to see how the replacement is happening<BR>\n" );
document.write( "3 + 5 + 7 +...+ (2k+1) + (2k+3) = k^2+4k+3<BR>\n" );
document.write( "<font color=red>3 + 5 + 7 +...+ (2k+1)</font> + (2k+3) = k^2+4k+3<BR>\n" );
document.write( "<font color=red>k(k+2) </font> + (2k+3) = k^2+4k+3\r<BR>\n" );
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document.write( "At this point, we just simplify the left side<BR>\n" );
document.write( "k(k+2) + (2k+3) = k^2+4k+3<BR>\n" );
document.write( "k^2+2k + 2k+3 = k^2+4k+3<BR>\n" );
document.write( "k^2+4k+3 = k^2+4k+3<BR>\n" );
document.write( "and we have another identity because the two sides are the same. <BR>\n" );
document.write( "We can be more rigorous and subtract x^2 from both sides, subtract 4x from both sides, and subtract 3 from both sides. Doing so will have us end up with 0 = 0 a much clearer identity we know to be always true; however this is more work than needed really.\r<BR>\n" );
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document.write( "So this means that if n = k holds true, then n=k+1 also holds true. This domino effect leads to proving the inductive step and proves the original claim to be true for all positive integers, or natural numbers.  </SPAN>\n" );
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