Use generating function to prove the identity
n
∑
k=o
r
k
s
n−k
=
r +s
n
"\\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"
"\\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 "x^2" -
"=[\\binom r0+\\binom r1x+\\binom r2x^2+....][\\binom s0+\\binom s1x+\\binom s2 x^2+...........]\n\n\n\\\\[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 -
"\\sum _{k=0}^n \\binom rk \\binom s{n-k}x^k"
and This is equated to "\\binom {n+m}{k} x^k."
Hence, "\\sum_{k=0}^n\\binom rk\\binom s{n-k}=\\binom {r+s} {n }"
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