document.write( "Question 252454: Determine the value of the product \r
\n" ); document.write( "\n" ); document.write( "(1- 1/2^2)(1-1/3^2)(1-1/4^2)....(1-1/99^2)(1-1/100^2)
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Algebra.Com's Answer #184421 by drk(1908)\"\" \"About 
You can put this solution on YOUR website!
We want the product of
\n" ); document.write( "(1-1/2^2)(1-1/3^2)(1-1/4^2)....(1-1/99^2)(1-1/100^2).\r
\n" ); document.write( "\n" ); document.write( "We can re-express this a the product of simple fractions as
\n" ); document.write( "(3/4)(8/9)(15/16)(24/25) . . . (9999/10000).
\n" ); document.write( "(3/4)(8/9) = 2/3 = [6/9]
\n" ); document.write( "(3/4)(8/9)(15/16) = 5/8 = [10/16]
\n" ); document.write( "(3/4)(8/9)(15/16)(24/25) = 3/5 = [15/25]
\n" ); document.write( "(3/4)(8/9)(15/16)(24/25)(35/36) = 7/12 = [21/36]
\n" ); document.write( ". . .
\n" ); document.write( "Notice the [parentheses] fractions.
\n" ); document.write( "There is a pattern to the numerator: 6, 10, 15, 21, . . .are all triangular numbers with the formula
\n" ); document.write( "\"n%5E2%2F2+%2B+3n%2F2+%2B+1\" where n>=2.
\n" ); document.write( "The denominator is just (n+1)^2, where n>=2.
\n" ); document.write( "Together, we get
\n" ); document.write( "\"n%5E2%2F2+%2B+3n%2F2+%2B+1\" / \"%28n%2B1%29%5E2\", n>=2.
\n" ); document.write( "I use two because the first multiplication took 2 fractions.
\n" ); document.write( "Now all of these together are 99 terms, so n = 99.
\n" ); document.write( "Using n = 99, we get
\n" ); document.write( "\"99%5E2%2F2+%2B3%2A99%2F2+%2B+1\" / \"100%5E2\" ~ .505\r
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