帕斯卡三角形 / Pascal's Triangle
你可以使用帕斯卡三角形来快速展开像 \((x+2y)^3\) 这样的表达式。
You can use Pascal's triangle to quickly expand expressions such as \((x+2y)^3\).
考虑 \((a+b)^n\) 对于 \(n = 0,1,2,3\) 和 4 的展开式：
Consider the expansions of \((a+b)^n\) for \(n = 0,1,2,3\) and 4:
\((a+b)^n\) 展开式中的每一项都有总指数 \(n\)：在 \(6a^2b^2\) 项中，总指数是 \(2+2=4\)。在 \(4ab^3\) 项中，总指数是 \(1+3=4\)。
Every term in the expansion of \((a+b)^n\) has total index \(n\): In the \(6a^2b^2\) term the total index is \(2+2=4\). In the \(4ab^3\) term the total index is \(1+3=4\).
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
展开式中的系数形成了一个被称为帕斯卡三角形的模式。
The coefficients in the expansions form a pattern that is known as Pascal's triangle.
帕斯卡三角形是通过将相邻的数字对相加来找到下一行的数字而形成的。
Pascal's triangle is formed by adding adjacent pairs of numbers to find the numbers on the next row.
帕斯卡三角形的性质 / Properties of Pascal's Triangle
帕斯卡三角形的第三行给出了 \((a+b)^2\) 展开式中的系数。
The third row of Pascal's triangle gives the coefficients in the expansion of \((a+b)^2\).
帕斯卡三角形的第 \((n+1)\) 行给出了 \((a+b)^n\) 展开式中的系数。
The \((n+1)\)th row of Pascal's triangle gives the coefficients in the expansion of \((a+b)^n\).
例题1 / Example 1
使用帕斯卡三角形求以下展开式：
Use Pascal's triangle to find the expansions of:
a) \((x+2y)^3\) b) \((2x-5)^4\)
a) \((x+2y)^3\) b) \((2x-5)^4\)
解答 / Solution
a) \((x+2y)^3\)
a) \((x+2y)^3\)
指数 = 3，所以看帕斯卡三角形的第4行来找到系数。
Index = 3 so look at the 4th row of Pascal's triangle to find the coefficients.
系数是 1, 3, 3, 1，所以：
The coefficients are 1, 3, 3, 1 so:
\((x+2y)^3 = 1 \cdot x^3 + 3 \cdot x^2(2y) + 3 \cdot x(2y)^2 + 1 \cdot (2y)^3\)
\((x+2y)^3 = 1 \cdot x^3 + 3 \cdot x^2(2y) + 3 \cdot x(2y)^2 + 1 \cdot (2y)^3\)
这是 \((a+b)^3\) 的展开式，其中 \(a = x\) 且 \(b = 2y\)。
This is the expansion of \((a+b)^3\) with \(a = x\) and \(b = 2y\).
\(= x^3 + 6x^2y + 12xy^2 + 8y^3\)
\(= x^3 + 6x^2y + 12xy^2 + 8y^3\)
使用括号确保你不会出错。\((2y)^2 = 4y^2\)。
Use brackets to make sure you don't make a mistake. \((2y)^2 = 4y^2\).
解答 / Solution
b) \((2x-5)^4\)
b) \((2x-5)^4\)
指数 = 4，所以看帕斯卡三角形的第5行。
Index = 4 so look at the 5th row of Pascal's triangle.
系数是 1, 4, 6, 4, 1，所以：
The coefficients are 1, 4, 6, 4, 1 so:
\((2x-5)^4 = 1 \cdot (2x)^4 + 4 \cdot (2x)^3(-5)^1 + 6 \cdot (2x)^2(-5)^2 + 4 \cdot (2x)^1(-5)^3 + 1 \cdot (-5)^4\)
\((2x-5)^4 = 1 \cdot (2x)^4 + 4 \cdot (2x)^3(-5)^1 + 6 \cdot (2x)^2(-5)^2 + 4 \cdot (2x)^1(-5)^3 + 1 \cdot (-5)^4\)
这是 \((a+b)^4\) 的展开式，其中 \(a = 2x\) 且 \(b = -5\)。
This is the expansion of \((a+b)^4\) with \(a = 2x\) and \(b = -5\).
\(= 16x^4 - 160x^3 + 600x^2 - 1000x + 625\)
\(= 16x^4 - 160x^3 + 600x^2 - 1000x + 625\)
注意负数时要小心。
Be careful with the negative numbers.
例题2 / Example 2
\((2-cx)^3\) 展开式中 \(x^2\) 的系数是 294。
The coefficient of \(x^2\) in the expansion of \((2-cx)^3\) is 294.
求常数 \(c\) 的可能值。
Find the possible value(s) of the constant \(c\).
解答 / Solution
系数是 1, 3, 3, 1：指数 = 3，所以使用帕斯卡三角形的第4行。
The coefficients are 1, 3, 3, 1: Index = 3 so use the 4th row of Pascal's triangle.
\(x^2\) 项是 \(3 \times 2(-cx)^2 = 6c^2x^2\)
The term in \(x^2\) is \(3 \times 2(-cx)^2 = 6c^2x^2\)
从 \((a+b)^3\) 的展开式中，\(x^2\) 项是 \(3ab^2\)，其中 \(a = 2\) 且 \(b = -cx\)。
From the expansion of \((a+b)^3\) the \(x^2\) term is \(3ab^2\) where \(a = 2\) and \(b = -cx\).
所以 \(6c^2 = 294\)
So \(6c^2 = 294\)
\(c^2 = 49\)
\(c^2 = 49\)
\(c = \pm 7\)
\(c = \pm 7\)
形成并求解关于 \(c\) 的方程。
Form and solve an equation in \(c\).