Binomial Theorem for Positive Integral Indices

Introduction to the Binomial Theorem

The Binomial Theorem provides a powerful formula for the expansion of any positive integral power of a binomial expression, such as $(x+a)^n$.

The Binomial Expansion for Positive Integral Index 'n'

For any positive integer n, the expansion is given by:

$$(x+a)^n = {^nC_0}x^n a^0 + {^nC_1}x^{n-1}a^1 + {^nC_2}x^{n-2}a^2 + ... + {^nC_n}x^0 a^n$$

Using summation notation, this can be written as:

$$(x+a)^n = \sum_{r=0}^{n} {^nC_r x^{n-r} a^r}$$

The coefficients $^nC_r$ are known as binomial coefficients.

Key Terms in the Expansion

• General Term: The $(r+1)^{th}$ term in the expansion is called the general term, denoted by $T_{r+1}$. It is crucial for finding any specific term without writing out the whole expansion.

$$T_{r+1} = {^nC_r}x^{n-r}a^r$$

• Term Independent of x: This is the term in which the power of x is zero (i.e., the constant term). To find it, we use the general term formula, set the exponent of x to 0, and solve for r.

• Number of Terms: The total number of terms in the expansion of $(x+a)^n$ is always $n+1$.

Example 1: Basic Expansion

Question: Expand $(x^2 + 2a)^4$ using the Binomial Theorem.

Solution: Here, n=4. The expansion is:

$$= {^4C_0}(x^2)^4(2a)^0 + {^4C_1}(x^2)^3(2a)^1 + {^4C_2}(x^2)^2(2a)^2 + {^4C_3}(x^2)^1(2a)^3 + {^4C_4}(x^2)^0(2a)^4$$

$$= 1(x^8)(1) + 4(x^6)(2a) + 6(x^4)(4a^2) + 4(x^2)(8a^3) + 1(1)(16a^4)$$

$$= x^8 + 8ax^6 + 24a^2x^4 + 32a^3x^2 + 16a^4$$

Example 2: Finding a Specific Term

Question: Find the 7th term in the expansion of $(4x - \frac{1}{2\sqrt{x}})^{13}$.

Solution: The 7th term corresponds to $T_{6+1}$, so $r=6$. Using the general term formula $T_{r+1} = {^nC_r}x^{n-r}a^r$ with $n=13, r=6$:

$$T_7 = {^{13}C_6}(4x)^{13-6}(-\frac{1}{2\sqrt{x}})^6 = {^{13}C_6}(4x)^7(\frac{1}{64x^3})$$

$$= {^{13}C_6} \cdot 4^7 \cdot x^7 \cdot \frac{1}{64} \cdot x^{-3} = {^{13}C_6} \cdot 16384 \cdot \frac{1}{64} \cdot x^4$$

$$= 1716 \cdot 256 \cdot x^4 = 439296x^4$$

Example 3: Finding the Term Independent of x

Question: Find the term independent of x in the expansion of $(\frac{3}{2}x^2 - \frac{1}{3x})^6$.

Solution: Let the $(r+1)^{th}$ term be independent of x. The general term is:

$$T_{r+1} = {^6C_r}(\frac{3}{2}x^2)^{6-r}(-\frac{1}{3x})^r = {^6C_r}(\frac{3}{2})^{6-r}(-1)^r(\frac{1}{3})^r (x^2)^{6-r}(x^{-1})^r$$

$$= {^6C_r}(\frac{3}{2})^{6-r}(-\frac{1}{3})^r x^{12-2r-r} = {^6C_r}(\frac{3}{2})^{6-r}(-\frac{1}{3})^r x^{12-3r}$$

For the term to be independent of x, the power of x must be 0: $12-3r=0 \implies r=4$. The required term is the $T_{4+1} = T_5$.

$$T_5 = {^6C_4}(\frac{3}{2})^{6-4}(-\frac{1}{3})^4 = {^6C_2}(\frac{3}{2})^2(\frac{1}{3^4}) = 15 \cdot \frac{9}{4} \cdot \frac{1}{81} = \frac{5}{12}$$