Answer to Question #192060 in Discrete Mathematics for ANJU JAYACHANDRAN

Question #192060

Use generating function to prove the identity

n

k=o



r

k

 s

n−k



=



r +s

n





1
Expert's answer
2021-05-13T01:18:41-0400

k=0n(rk)(snk)=(r+sn)\sum_{k=0}^n\binom rk\binom s{n-k}=\binom {r+s} {n }


we know, (1+x)r+s=(1+x)r(1+x)s(1+x)^{r+s}=(1+x)^r(1+x)^s


k=0r+s(r+sn)xn=j=0r(rj)xj+a=os(sa)xa\sum_{k=0}^{r+s}\binom {r+s}{n}x^n=\sum_{j=0}^r \binom{r}{j}x^j+\sum_{a=o}^s\binom{s}{a}x^a


Now we expand each of the summation on RHS upto x2x^2 -


=[(r0)+(r1)x+(r2)x2+....][(s0)+(s1)x+(s2)x2+...........]=[(r0)(s0)]+[(r0)(s1)+[(s0)(r1)]x+[(r1)(s1)+(r0)(s2)+(r2)(s0)]x2.......=[\binom r0+\binom r1x+\binom r2x^2+....][\binom s0+\binom s1x+\binom s2 x^2+...........] \\[9pt]=[\binom r0\binom s0]+[\binom r0\binom s1+[\binom s0\binom r1]x+[\binom r1\binom s1+\binom r0\binom s2+\binom r2\binom s0]x^2.......


Thus By inspecting a clear Pattern is evident and the n^{th} Power of ''x'' is of the form -


k=0n(rk)(snk)xk\sum _{k=0}^n \binom rk \binom s{n-k}x^k


and This is equated to (n+mk)xk.\binom {n+m}{k} x^k.


Hence, k=0n(rk)(snk)=(r+sn)\sum_{k=0}^n\binom rk\binom s{n-k}=\binom {r+s} {n }


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