Recall the binomial theorem:
[tex]\displaystyle (a+b)^n = \sum_{k=0}^n \binom nk a^{n-k} b^k[/tex]
a. Let a = b = 1 and n = 8. Then the sum above gives the sum shown here,
[tex]\displaystyle \sum_{k=0}^8 \binom8k 1^{n-k} 1^k = \sum_{k=0}^8 \binom8k[/tex]
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
[tex]\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[/tex]
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.
