document.write( "Question 1089122: In the expansion of (1+x)^21,the coefficient of (2r+1)th term is equal to the coefficient of (3r+2)th term, then find the value of r.
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Algebra.Com's Answer #703489 by KMST(5328)\"\" \"About 
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The expansion of a binomial \"%28a%2Bb%29%5En\" has \"n%2B1\" terms.
\n" ); document.write( "They are of the form \"Ca%5Epb%5Eq\" with a combinatorial coefficient \"C\" ,
\n" ); document.write( "and positive integers \"p\" and \"q\" such that \"p%2Bq=n\" .
\n" ); document.write( "There's is a certain symmetry to the combinatorial coefficients:
\n" ); document.write( "The first and last ones (the number \"1\" and \"n%2B1\" coefficients are 1;
\n" ); document.write( "the number \"2\" and \"n\" coefficients are \"n\" ;
\n" ); document.write( "the number \"3\" and \"n-1\" coefficients are \"n%28n-1%29%2F2\" ,
\n" ); document.write( "and so on, so that the number \"m\" and \"s\" coefficients will be the same if \"m%2Bs=n%2B2\" .
\n" ); document.write( "So, in this case,
\n" ); document.write( "\"%282r%2B1%29%2B%283r%2B2%29=21%2B2\" ,
\n" ); document.write( "\"5r%2B3=23\" ,
\n" ); document.write( "\"5r=20\" ,
\n" ); document.write( "and \"highlight%28r=4%29\" .
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\n" ); document.write( "Verification:
\n" ); document.write( "\"2r%2B1=2%2A4%2B1=8%2B1=9\"
\n" ); document.write( "\"3r%2B2=3%2A4%2B2=12%2B2=14\"
\n" ); document.write( "The 9th coefficient is
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\n" ); document.write( "The 14th coefficient is
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