更多详细例题
每个展开式都使用 \((a+b)^n = \sum_{k=0}^{n}\binom{n}{k}\,a^{\,n-k}\,b^{\,k}\)，并直接从帕斯卡三角形的相应行中获取其系数。
例1：\((x-2)^4\) — 符号交替
这里 \(a = x\)，\(b = -2\)，\(n = 4\)。帕斯卡三角形的第4行是 \(1, 4, 6, 4, 1\)。因为 \(b\) 是负数，\(-2\) 的幂次使符号交替：
\(\binom{4}{0}x^4(-2)^0 = 1\cdot x^4 \cdot 1 = x^4\)
\(\binom{4}{1}x^3(-2)^1 = 4\cdot x^3 \cdot(-2) = -8x^3\)
\(\binom{4}{2}x^2(-2)^2 = 6\cdot x^2 \cdot 4 = 24x^2\)
\(\binom{4}{3}x^1(-2)^3 = 4\cdot x \cdot(-8) = -32x\)
\(\binom{4}{4}x^0(-2)^4 = 1\cdot 1 \cdot 16 = 16\)
化简：\((x-2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16\)。
例2：\((2+3)^5\) — 完全数值
这里 \(a = 2\)，\(b = 3\)，\(n = 5\)，第5行是 \(1, 5, 10, 10, 5, 1\)：
\(\binom{5}{0}2^5 3^0 = 1\cdot 32\cdot 1 = 32\)
\(\binom{5}{1}2^4 3^1 = 5\cdot 16\cdot 3 = 240\)
\(\binom{5}{2}2^3 3^2 = 10\cdot 8\cdot 9 = 720\)
\(\binom{5}{3}2^2 3^3 = 10\cdot 4\cdot 27 = 1080\)
\(\binom{5}{4}2^1 3^4 = 5\cdot 2\cdot 81 = 810\)
\(\binom{5}{5}2^0 3^5 = 1\cdot 1\cdot 243 = 243\)
求和各项：\(32 + 240 + 720 + 1080 + 810 + 243 = \) 3125。作为检验，\((2+3)^5 = 5^5 = 3125\)。
例3：\(\left(1+\tfrac{1}{2}\right)^3\) — 分数底数
这里 \(a = 1\)，\(b = \tfrac{1}{2}\)，\(n = 3\)，第3行等于 \(1, 3, 3, 1\)：
\(\binom{3}{0}1^3\left(\tfrac{1}{2}\right)^0 = 1\cdot 1\cdot 1 = 1\)
\(\binom{3}{1}1^2\left(\tfrac{1}{2}\right)^1 = 3\cdot 1\cdot\tfrac{1}{2} = \tfrac{3}{2}\)
\(\binom{3}{2}1^1\left(\tfrac{1}{2}\right)^2 = 3\cdot 1\cdot\tfrac{1}{4} = \tfrac{3}{4}\)
\(\binom{3}{3}1^0\left(\tfrac{1}{2}\right)^3 = 1\cdot 1\cdot\tfrac{1}{8} = \tfrac{1}{8}\)
将各项相加：\(1 + \tfrac{3}{2} + \tfrac{3}{4} + \tfrac{1}{8} = \tfrac{8+12+6+1}{8} = \tfrac{27}{8} = \) 3.375。这与 \(\left(\tfrac{3}{2}\right)^3 = \tfrac{27}{8}\) 相符。