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I hate asking this because I'm usually extremely good at these but I'm STUCK on two (this is the first time this semester in this class)...
1. If n is a positive integer Verify by substitution that: (2n choose n+1) + (2n choose n) = ½(2n+2 choose n+1). Then count combinatorially.
2. Also Need to prove that Sum(k from 0 to n) (
choose k)( choose n-k )= (+ choose n).I realize this is a lot to ask. So I thank anyone who can help.
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Actually... for #1 (the verify by substitution part) is this correct:
n choose r = n! / [r!(n-r)!]
Applied to left side:
2n choose (n+1) = (2n)! / [(n+1)! (2n-(n+1)!]
= (2n)! / [(n+1)!(n-1)!]
2n choose n = (2n)! / [(n!(2n-n)!]
= (2n)! / [n!n!]
Then
(2n)! / [(n+1)!(n-1)!] + (2n)! / [n!n!]
= (2n)! {1/[(n+1)!(n-1)!] + 1/(n!n!)}
The common denominator will be (n+1)!n!
= (2n)! {n/[(n+1)!n!] + (n+1)/[(n+1)!n!]
= (2n)! {(n+n+1) / [(n+1)!n!]}
= (2n)! {(2n+1) / [(n+1)!n!]}
= (2n+1)! / [(n+1)!n!]
Multiply the top and bottom by (2n+2)
= (2n+2)! / [(2n+2)(n+1)!n!]}
= (2n+2)! / [2(n+1)(n+1)!n!]
= (1/2)(2n+2)! / [(n+1)!(n+1)!]
= (1/2) (2n+2 choose n+1)
?
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