2. Show that \left(1+x^2+2x\right)^{2n}=\sum _{k=0}^n\binom{2n}{k}x^{2n-k}\left(x+2\right)^{2n-k}.\: 3. It is known that x^{2n-k}\left(x+2\right)^{2n-k}=\binom{2n-k}{0}2^{2n-k}x^{2n-k}+\binom{2n-k}{1}2^{2n-k-1}x^{2n-k+1}+...+\binom{2n-k}{2n-k}2^{0}x^{4n-2k}. Show that \binom{4n}{2n}=\sum _{k=0}^n2^{2n-2k}\binom{2n}{k}\binom{2n-k}{k}\: I'm stuck from question 2. (question one was writing coefficient of x^2n in binomial expansion of (1+x)^4n) Any tips/hints? Thanks for your help :D!
^{2n}=\sum _{k=0}^n\binom{2n}{k}x^{2n-k}\left(x+2\right)^{2n-k}.\:)
^{2n-k}=\binom{2n-k}{0}2^{2n-k}x^{2n-k}+\binom{2n-k}{1}2^{2n-k-1}x^{2n-k+1}+...+\binom{2n-k}{2n-k}2^{0}x^{4n-2k}.)
