Maybe can someone help? Is the equation with n ≥ 0 correct? (easier to solve by using generating function)
\(\displaystyle \displaystyle \mbox{2. }\quad \sum_{0\leq k \leq n/2}\, (-1)^k\, \binom{n\, -\, k}{k}\, 2^{n-2k}\, =\, n\, +\, 1\quad ?\)
\(\displaystyle \displaystyle \mbox{3. }\quad \sum_{k=0}^{n}\, \binom{n\, +\, k}{2k}\, 2^{n-k}\, =\, \dfrac{2^{2n+1}\, +\, 1}{3}\quad ?\)
\(\displaystyle \displaystyle \mbox{4. }\quad \sum_{k=0}^n\, \binom{2k}{k}\, \binom{n}{k}\, \left(-\dfrac{1}{4}\right)^k\, =\, 2^{-2n}\, \binom{2n}{n}\quad ?\)
\(\displaystyle \displaystyle \mbox{5. }\quad \sum_k\, \binom{m}{k}\, \binom{n\, +\, k}{m}\, =\, \sum_k\, \binom{m}{k}\, \binom{n}{k}\, 2^k\quad ?\)
. . . . .\(\displaystyle \mbox{Here }\, \binom{n}{k}\, = 0\, \mbox{ if }\, k\, >\, n.\)
\(\displaystyle \displaystyle \mbox{2. }\quad \sum_{0\leq k \leq n/2}\, (-1)^k\, \binom{n\, -\, k}{k}\, 2^{n-2k}\, =\, n\, +\, 1\quad ?\)
\(\displaystyle \displaystyle \mbox{3. }\quad \sum_{k=0}^{n}\, \binom{n\, +\, k}{2k}\, 2^{n-k}\, =\, \dfrac{2^{2n+1}\, +\, 1}{3}\quad ?\)
\(\displaystyle \displaystyle \mbox{4. }\quad \sum_{k=0}^n\, \binom{2k}{k}\, \binom{n}{k}\, \left(-\dfrac{1}{4}\right)^k\, =\, 2^{-2n}\, \binom{2n}{n}\quad ?\)
\(\displaystyle \displaystyle \mbox{5. }\quad \sum_k\, \binom{m}{k}\, \binom{n\, +\, k}{m}\, =\, \sum_k\, \binom{m}{k}\, \binom{n}{k}\, 2^k\quad ?\)
. . . . .\(\displaystyle \mbox{Here }\, \binom{n}{k}\, = 0\, \mbox{ if }\, k\, >\, n.\)
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