document.write( "Question 945137: an equilateral triangle is drawn by joining the midpoints of the sides of another equilateral triangle. a third equilateral triangle is drawn inside the second one joining the midpoints of the sides of the second equilateral triangle, and the process is continued indefinitely. find the sum of all perimeters of all equilateral triangles, if the side of the largest equilateral triangle is 24? \n" ); document.write( "
Algebra.Com's Answer #576513 by KMST(5328)\"\" \"About 
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If the side of the first/largest equilateral triangle is 24,
\n" ); document.write( "its perimeter is \"24%2A3=72\" .
\n" ); document.write( "Joining the midpoints of the sides of an equilateral triangle
\n" ); document.write( " splits that triangle into
\n" ); document.write( "4 equilateral triangles whose sides are half as long as the sides of the original equilateral triangle.
\n" ); document.write( "Consequently, the perimeter of the second triangle will be half the perimeter of the first.
\n" ); document.write( "So, the sum of all the infinite perimeters is
\n" ); document.write( " .
\n" ); document.write( "The sum \"1%2B1%2F2%2B1%2F4%2B%22...%22\" is the sum of an infinite geometric sequence,
\n" ); document.write( "with first term \"b%5B1%5D=1\" and common ratio \"r=1%2F2\" .
\n" ); document.write( "It is easy to see that it adds up to \"2\" , because
\n" ); document.write( "when you added \"1\" more term to the first term, you were \"1%2F2\" short of \"2\" ;
\n" ); document.write( "when you added to that term number \"2\" terms you were \"1%2F4=1%2F2%5E2\" short of \"2\" ,
\n" ); document.write( "and you keep being \"1%2F2%5En\" short of \"2\" as \"n\" keeps increasing.
\n" ); document.write( "If you (or your teacher) insist on using formulas,
\n" ); document.write( "the sum of the first \"n\" terms of geometric progression (if such a sum exists) is
\n" ); document.write( "\"b%5B1%5D%2A%28%281-r%5En%29%2F%281-r%29%29\" , and with \"r%3C1\" ,
\n" ); document.write( "the sum of the infinite terms of geometric progression is
\n" ); document.write( "\"b%5B1%5D%2A%281%2F%281-r%29%29\" .
\n" ); document.write( "In this case \n" ); document.write( "
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