document.write( "Question 1045685: prove that C(n,r) + C(n-1,r) + C(n-2, r) +.......+ C(r,r) = C(n+1,r+1) \n" ); document.write( "

Algebra.Com's Answer #661291 by robertb(5830) $\"\"$ $\"About$

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Proceed by induction: Consider the expression $\"C%28r%2Bk%2Cr%29+%2B+C%28r%2Bk-1%2Cr%29\"$ + ... + $\"+C%28r%2B1%2Cr%29+%2B+C%28r%2Cr%29\"$ \r
\n" ); document.write( "\n" ); document.write( "For k = 1: $\"C%28r%2B1%2Cr%29+%2B+C%28r%2Cr%29+=+%28r%2B1%29+%2B1+=+r%2B2+=+C%28r%2B2%2Cr%2B1%29\"$ , and the relation is true.\r
\n" ); document.write( "\n" ); document.write( "Suppose the relation is also true up to k = n - r, so that\r
\n" ); document.write( "\n" ); document.write( " $\"C%28n%2Cr%29+%2B+C%28n-1%2Cr%29\"$ + ... + $\"C%28r%2B2%2Cr%29+%2B+C%28r%2B1%2Cr%29+%2B+C%28r%2Cr%29+=+C%28n%2B1%2Cr%2B1%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "===> $\"C%28n%2B1%2Cr%29+%2B+C%28n%2Cr%29+%2B+C%28n-1%2Cr%29\"$ + ... + $\"+C%28r%2B2%2Cr%29+%2B+C%28r%2B1%2Cr%29+%2B+C%28r%2Cr%29+\"$ \r
\n" ); document.write( "\n" ); document.write( "= $\"+C%28n%2B1%2Cr%29+%2B+C%28n%2B1%2Cr%2B1%29\"$ \r
\n" ); document.write( "\n" ); document.write( "= $\"C%28n%2B2%2C+r%2B1%29\"$ , by a direct application of Pascal's Triangle $\"C%28N%2CR%29+%2B+C%28N%2CR-1%29+=+C%28N%2B1%2CR%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "Hence $\"C%28n%2B1%2Cr%29+%2B+C%28n%2Cr%29+%2B+C%28n-1%2Cr%29+\"$ + ... + $\"C%28r%2B2%2Cr%29+%2B+C%28r%2B1%2Cr%29+%2B+C%28r%2Cr%29+%0D%0A%0D%0A=+C%28n%2B2%2C+r%2B1%29\"$ , \r
\n" ); document.write( "\n" ); document.write( "and the relation is proved. \n" ); document.write( "

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