( C 0 + C 1 ) ( C 1 + C 2 ) ( C 2 + C 3 ) ⋅ ⋅ ⋅ ( C n − 1 + C n )
Let C k denote the binomial coefficient ( k n ) , can you simplify the product above for positive integer n > 1 ?
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Can you elaborate on it?
On using the property C r n + C r − 1 n = C r n + 1 it gets reduced to the form C 1 n + 1 C 2 n + 1 C 3 n + 1 . . . . . C n n + 1
On simplification, we get
C 1 n + 1 = 1 ! n ! ( n + 1 ) ! = ( n + 1 )
C 2 n + 1 = 2 ! ( n − 1 ) ! ( n + 1 ) ! = 2 n ( n + 1 ) = 2 C 1 n ( n + 1 )
C 3 n + 1 = 3 ! ( n − 2 ) ! ( n + 1 ) ! = 3 ∗ 2 n ( n + 1 ) ( n − 1 ) = 3 C 2 n ( n + 1 ) and so on...
Since C n = 1 , the first and second option are technically equal.
The formula that I derived is a little bit different but serves the same purpose in the end..
nCr + nC(r+1)=(n+1)C(r+1).
But not recommended for this problem..gets too hairy in the later part.lol.but still a good result to be remembered..
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Use the property C r n + C r − 1 n = C r n + 1