document.write( "Question 71405: sum the series\r
\n" ); document.write( "\n" ); document.write( "1+4/6 + 4.5/6.9 + 4.5.6/6.9.12 + ................infinity \n" ); document.write( "
Algebra.Com's Answer #583207 by hafys14feb(1)\"\" \"About 
You can put this solution on YOUR website!
1+ 4/6+ 4.5/6.9+ 4.5.6/6.9.12+ ...infinity
\n" ); document.write( "Let S= 1+ 4/6+ 4.5/6.9+ 4.5.6/6.9.12+ ...infinity
\n" ); document.write( " = 1+ 4/2.3 + 4.5/2.3.3.3 + 4.5.6/2.3.3.3.3.4 +...
\n" ); document.write( " = 1+ 4/1.2 [1/3] + 4.5/1.2.3 [1/3]^2 + 4.5.6/1.2.3.4 [1/3]^3 +...
\n" ); document.write( " Inserting the missing factor in the numerator
\n" ); document.write( "Multiply both sides by 3 ....(4,5,6 ...are in Ap.the preceding term is 3) \r
\n" ); document.write( "\n" ); document.write( " 3S =3+ 3.4/2![1/3] +3.4.5/3! [1/3]^2 +3.4.5.6/4! [1/3]^3+...\r
\n" ); document.write( "\n" ); document.write( "Now Multiply both sides by [1/3]..\r
\n" ); document.write( "\n" ); document.write( "3S= 3/1! [1/3]+ 3.4/2! [1/3]^2 + 3.4.5/3! [1/3]^3 + ...
\n" ); document.write( "Now add 1 on both sides ;
\n" ); document.write( " S+1=1+3[1/3]+ 3.4/2![1/3]^2 + 3.4.5/3![1/3]^3 +...
\n" ); document.write( "This is in the form of
\n" ); document.write( "1+p/1![x/q] +p(p+q)/2![x/q]^2+p(p+q)(p+2q)/3! [x/q]^3...
\n" ); document.write( "which is equal to (1-x)^-p/q\r
\n" ); document.write( "\n" ); document.write( "So ,
\n" ); document.write( "S+1= (1-x)^-p/q ______equation 1
\n" ); document.write( "here p=3 and p+q=4
\n" ); document.write( "Therefore q=4-3=1
\n" ); document.write( "and we know x/q=1/3
\n" ); document.write( "so x= 1/3.\r
\n" ); document.write( "\n" ); document.write( "Again by solving the equation 1,\r
\n" ); document.write( "\n" ); document.write( "S+1= (1-1/3)^-3/1
\n" ); document.write( " =(2/3)^-3
\n" ); document.write( " S = 19/8
\n" ); document.write( "The sum is 19/8.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );