Question 48
Prove that
$$
2 \sum_{k=0}^{n} \binom{2n}{k} = \binom{2n}{n} + 2^{2n}, \ n = 1, 2, \dots
$$
Prove also that
$$
2 \sum_{k=0}^{n} \binom{2n}{k} (-1)^k = \binom{2n}{n} (-1)^n, \ n = 1, 2, \dots
$$
$$
2 \sum_{k=0}^{n} \binom{2n}{k} = \binom{2n}{n} + 2^{2n}, \ n = 1, 2, \dots
$$
Prove also that
$$
2 \sum_{k=0}^{n} \binom{2n}{k} (-1)^k = \binom{2n}{n} (-1)^n, \ n = 1, 2, \dots
$$