The binomial theorem says
[tex]\displaystyle (a + b)^n = \sum_{k=0}^n \binom nk a^{n-k} b^k \\\\(a+b)^n= \binom n0 a^n + \binom n1 a^{n-1}b + \cdots + \binom n{n-1} ab^{n-1} + \binom nn b^n[/tex]
where
[tex]\dbinom nk = \dfrac{n!}{k!(n-k)!}[/tex]
is the so-called binomial coefficient. The binomial coefficients follow a neat pattern called Pascal's triangle:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
where the n-th row, starting with n = 0, lists the coefficient of the k-th term in the sum (with 0 ≤ k ≤ n). For example,
[tex]\displaystyle n=0 \implies (a + b)^0 = \sum_{k=0}^0 \binom nk a^{0-k} b^k = \binom00 a^{0-0} b^0 = \underline{1}[/tex]
[tex]\displaystyle n=1 \implies (a + b)^1 = \sum_{k=0}^1 \binom 1k a^{1-k} b^k = \binom10 a^{1-0} b^0 + \binom11 a^{1-1}b^1 = \underline{1}a + \underline{1}b[/tex]
[tex]\displaystyle n=2 \implies (a + b)^2 = \sum_{k=0}^2 \binom 2k a^{2-k} b^k = \binom20a^{2-0}b^0+\binom21a^{2-1}b^1+\binom22a^{2-2}b^2 = \underline{1}a^2 + \underline{2}ab+\underline{1}b^2[/tex]
Each row starts and ends with 1, and every coefficient in between is obtained by adding the coefficients directly above and to the left. The next row, for instance, would be
1 1 + 4 4 + 6 6 + 4 4 + 1 1
or
1 5 10 10 5 1
which is to say,
[tex]\displaystyle (a + b)^5 = \underline{1}a^5+\underline{5}a^4b+\underline{10}a^3b^2+\underline{10}a^2b^3+\underline{5}ab^4+\underline{1}b^5[/tex]
In your case, we have (d - 4b)², so we take a, b, and n above with d, -4b, and 2, respectively:
[tex]\displaystyle (d - 4b)^2 = \sum_{k=0}^2 \binom 2k d^{2-k} (-4b)^k[/tex]
[tex]\displaystyle (d - 4b)^2 = \sum_{k=0}^2 \binom 2k (-4)^k d^{2-k} b^k[/tex]
[tex]\displaystyle (d - 4b)^2 = \binom20 (-4)^0 d^{2-0} b^0 + \binom21 (-4)^1 d^{2-1} b^1 + \binom22 (-4)^2 d^{2-2} b^2[/tex]
[tex]\displaystyle (d - 4b)^2 = 1\cdot1\cdot d^2\cdot1 + 2 \cdot (-4) d \cdot b + 1 \cdot 16 \cdot 1 \cdot b^2[/tex]
[tex]\boxed{(d - 4b)^2 = d^2 - 8bd + 16b^2}[/tex]
