1. Meaning of the General Term
In the expansion of \((a + b)^n\), every term follows a predictable pattern. Instead of writing the entire expansion, we can directly find any term using a formula called the general term. This lets us locate terms in the middle or near the end without expanding everything.
2. Formula for the General Term
The binomial theorem gives a direct formula for the \((r+1)\)-th term of \((a + b)^n\).
2.1. General Term Formula
The formula for the \((r+1)\)-th term is:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^{r} \]
Here, \( r = 0, 1, 2, \ldots, n \). The powers of \(a\) decrease while the powers of \(b\) increase.
2.2. Pattern in Terms
The first few terms obtained from the formula are:
- \(T_1 = \binom{n}{0} a^n b^0\)
- \(T_2 = \binom{n}{1} a^{n-1} b^1\)
- \(T_3 = \binom{n}{2} a^{n-2} b^2\)
Each term follows the same format.
3. How to Use the Formula
To find a specific term in the expansion, decide which term number you want and substitute the value into the formula.
3.1. Step-by-Step Method
- Identify the power \(n\) in \((a + b)^n\).
- Find which term you want—say the \(k\)-th term.
- Use \( r = k - 1 \).
- Substitute into \[ T_{r+1} = \binom{n}{r} a^{n-r} b^{r}. \]
4. Worked Examples
These examples show how to apply the formula quickly without expanding everything.
4.1. Example 1: Find the 5th term of \((x + 2)^7\)
Here, \(n = 7\). For the 5th term:
\(r = 5 - 1 = 4\)
\[T_5 = \binom{7}{4} x^{7-4} 2^4\]
- \(\binom{7}{4} = 35\)
- \(x^3\)
- \(2^4 = 16\)
So, \(T_5 = 35 \cdot 16 \cdot x^3 = 560x^3\).
4.2. Example 2: Find the 3rd term of \((3a - b)^8\)
For the 3rd term:
\(r = 3 - 1 = 2\)
\[T_3 = \binom{8}{2} (3a)^{6} (-b)^2\]
- \(\binom{8}{2} = 28\)
- \((3a)^6 = 729a^6\)
- \((-b)^2 = b^2\)
So, \(T_3 = 28 \cdot 729a^6 b^2 = 20412a^6 b^2\).