document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=986170>Question 986170</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Jack is building a pyramid by leaning cards against each other. He starts with 2 cards. This is a \"1-level\" pyramid and it uses 2 cards. <BR>\n" );
document.write( "By making the base wider, and putting a 'floor' between each level, he can make higher pyramids... A '2 level' pyramid uses 7 cards, a '3 level' pyramid uses 15 cards and so on. <BR>\n" );
document.write( "Write an equation showing how many cards are needed (c) for an 'n-level' pyramid  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #606997 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=606997\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Draw out a few cases. I drew out pyramids with 1 level, 2 levels, 3 levels, 4 levels, and 5 levels. Basically I am showing below cases n = 1 through n = 5 \r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width=\"400px\" src = \"http://i150.photobucket.com/albums/s91/jim_thompson5910/cards2_zps7mmzrum8.png\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The list goes on, but eventually we run out of room, so it's a good place to stop here. We have enough to see a pattern forming.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Notice the sequence of cards needed: 2, 7, 15, 26, 40, ...\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The jump from 2 to 7 is +5<BR>\n" );
document.write( "The jump from 7 to 15 is +8<BR>\n" );
document.write( "The jump from 15 to 26 is +11<BR>\n" );
document.write( "The jump from 26 to 40 is +14\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "We have a list of common differences: 5, 8, 11, 14<BR>\n" );
document.write( "Each common difference is separated by 3. <BR>\n" );
document.write( "5+3 = 8<BR>\n" );
document.write( "8+3 = 11<BR>\n" );
document.write( "11+3 = 14<BR>\n" );
document.write( "and so on...\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Since we have constant 2nd differences (all 3), this means a quadratic polynomial models the original sequence.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "----------------------------------------------------------\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Pair up this list {1,2,3,4,5} with {2, 7, 15, 26, 40} to get this list of ordered pairs\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "(1,2), (2,7), (3,15), (4,26), (5,40)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "we only need 3 ordered pairs. Let's just work with the first three ordered pairs. We will work with the general quadratic form y = ax^2+bx+c\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Plug in (x,y) = (1,2) into y = ax^2+bx+c to get<BR>\n" );
document.write( "y = ax^2+bx+c<BR>\n" );
document.write( "2 = a*1^2+b*1+c<BR>\n" );
document.write( "a+b+c = 2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Plug in (x,y) = (2,7) into y = ax^2+bx+c to get<BR>\n" );
document.write( "y = ax^2+bx+c<BR>\n" );
document.write( "7 = a*2^2+b*2+c<BR>\n" );
document.write( "4a+2b+c = 7\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Plug in (x,y) = (3,15) into y = ax^2+bx+c to get<BR>\n" );
document.write( "y = ax^2+bx+c<BR>\n" );
document.write( "15 = a*3^2+b*3+c<BR>\n" );
document.write( "9a+3b+c = 15\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "-------------------------------------------------------\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "After plugging in those (x,y) pairs and simplifying, we have this system of equations\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "a+b+c = 2<BR>\n" );
document.write( "4a+2b+c = 7<BR>\n" );
document.write( "9a+3b+c = 15\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "There are a number of ways to solve. I'm going to use a calculator. Specifically the TI83's inverse matrix feature.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Doing so essentially solves for (a,b,c) simultaneously. The calculator states that\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "a = 1.5<BR>\n" );
document.write( "b = 0.5<BR>\n" );
document.write( "c = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So the nth term equation is <font size=5 color=\"red\">c = 1.5n^2 + 0.5n</font> where n is the level number and c is the number of cards needed to make the pyramid that tall.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "-------------------------------------------------------\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "For instance, if n = 3, then we should get c = 15. Let's see if that's true.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "c = 1.5n^2 + 0.5n<BR>\n" );
document.write( "c = 1.5*3^2 + 0.5*3<BR>\n" );
document.write( "c = 1.5*9 + 0.5*3<BR>\n" );
document.write( "c = 13.5 + 1.5<BR>\n" );
document.write( "c = 15\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "we definitely get c = 15. So we need 15 cards for a 3 level pyramid. This confirms the equation is true for n = 3. Try other values of n (like n = 1 or n = 2) to generate other values of c. This is a good way to confirm we have the correct nth term formula.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );