Recall the binomial theorem:
![\displaystyle (a+b)^n = \sum_(k=0)^n \binom nk a^(n-k) b^k](https://img.qammunity.org/2022/formulas/mathematics/college/ack8iqk2n2srash6nga0o3ds22dzqd5b2a.png)
a. Let a = b = 1 and n = 8. Then the sum above gives the sum shown here,
![\displaystyle \sum_(k=0)^8 \binom8k 1^(n-k) 1^k = \sum_(k=0)^8 \binom8k](https://img.qammunity.org/2022/formulas/mathematics/college/d9he4t7w5jxtbvu7dm18wszfnh2drk70ie.png)
so it reduces to (1 + 1)⁸ = 2⁸ = 256.
b. Let a = 1 and b = -1 and k = 8. Then by the same argument, the sum reduces to (1 - 1)⁸ = 0.
c. Notice that
![\displaystyle \sum_(j=0)^n \binom n{n-j} = \binom nn + \binom n{n-1} + \binom n{n-2} + \cdots \binom n2 + \binom n1 + \binom n0](https://img.qammunity.org/2022/formulas/mathematics/college/cgw9t92gabhu2bhclqv1hjl73zg3npwg04.png)
but this is nearly identical to the sum in part a, with a = b = 1 and arbitrary n, but the order of terms is reversed. Then it reduces to (1 + 1)ⁿ = 2ⁿ.
d. Let a = 1 and b = 2. Then the sum is (1 + 2)ⁿ = 3ⁿ.