Example 1: Find the 5th term in the expansion of \( (3x - 2)^7 \) (Medium)
Solution:
\[ T_5 = \binom{7}{4} (3x)^{7-4} (-2)^4 = \binom{7}{4} (3x)^3 (16) \] \[ = 35 \times 27 x^3 \times 16 = 35 \times 432 x^3 = 15120 x^3 \]Answer: \( 15120 x^3 \)
The Binomial Theorem is a powerful tool in algebra that allows us to expand expressions of the form \( (a + b)^n \) where \( n \) is a non-negative integer. Beyond expansion, it has numerous applications including finding particular terms, approximations for expressions with small terms, and determining the greatest coefficient in the expansion.
The binomial expansion of \( (a + b)^n \) is given by:
\[ (a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r \]Here, the general term (also called the \( (r+1)^{th} \) term) is:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \quad \text{for } r = 0, 1, 2, \ldots, n \]This formula is crucial for finding any specific term without expanding the entire expression.
Middle Term: When \( n \) is even, the middle term is the \( \left(\frac{n}{2} + 1\right)^{th} \) term. When \( n \) is odd, there are two middle terms: the \( \left(\frac{n+1}{2}\right)^{th} \) and \( \left(\frac{n+3}{2}\right)^{th} \) terms.
Example: Find the 4th term in the expansion of \( (2x - 3)^5 \).
Solution:
\[T_4 = \binom{5}{3} (2x)^{5-3} (-3)^3 = \binom{5}{3} (2x)^2 (-27) = 10 \times 4x^2 \times (-27) = -1080 x^2\]The binomial theorem is also used to approximate expressions of the form \( (1 + x)^n \) when \( |x| < 1 \) and \( n \) is any real number (not necessarily an integer). For small \( x \), higher powers of \( x \) become very small and can be ignored for approximation.
For example, for \( |x| \ll 1 \),
\[(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2} x^2 + \ldots\]This is particularly useful in physics and engineering for simplifying complex expressions.
Example: Approximate \( (1.02)^5 \) using the binomial theorem.
Solution:
\[(1 + 0.02)^5 \approx 1 + 5 \times 0.02 + \frac{5 \times 4}{2} (0.02)^2 = 1 + 0.1 + 10 \times 0.0004 = 1 + 0.1 + 0.004 = 1.104\]The exact value is approximately 1.10408, so the approximation is quite close.
In the expansion of \( (a + b)^n \), the binomial coefficients \( \binom{n}{r} \) determine the coefficients of each term. The greatest coefficient is the maximum value among these binomial coefficients.
Properties:
Finding the Greatest Coefficient:
For \( n \) even, the greatest coefficient is \( \binom{n}{\frac{n}{2}} \).
For \( n \) odd, the two middle coefficients \( \binom{n}{\frac{n-1}{2}} \) and \( \binom{n}{\frac{n+1}{2}} \) are equal and greatest.
Example: Find the greatest coefficient in the expansion of \( (1 + x)^{10} \).
Solution:
\[\text{Greatest coefficient} = \binom{10}{5} = \frac{10!}{5!5!} = 252\]The binomial theorem is widely used in probability theory, combinatorics, and algebraic identities. For example, the coefficients \( \binom{n}{r} \) represent the number of ways to choose \( r \) elements from \( n \), which is fundamental in counting problems.
In probability, the binomial distribution uses these coefficients to calculate probabilities of exactly \( r \) successes in \( n \) independent Bernoulli trials.
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
Pascal's Triangle shows binomial coefficients visually, illustrating symmetry and the pattern of coefficients.
Solution:
\[ T_5 = \binom{7}{4} (3x)^{7-4} (-2)^4 = \binom{7}{4} (3x)^3 (16) \] \[ = 35 \times 27 x^3 \times 16 = 35 \times 432 x^3 = 15120 x^3 \]Answer: \( 15120 x^3 \)
Solution:
\[ \sqrt[3]{1.05} = (1 + 0.05)^{\frac{1}{3}} \approx 1 + \frac{1}{3}(0.05) - \frac{1}{9} \frac{(0.05)^2}{2} = 1 + 0.0167 - 0.00139 = 1.0153 \]Answer: Approximately \( 1.0153 \)
Solution:
\[ \text{Greatest coefficient} = \binom{12}{6} = \frac{12!}{6!6!} = 924 \]Answer: 924
Solution:
General term:
\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} \left(\frac{1}{x}\right)^r = \binom{6}{r} x^{2(6-r)} x^{-r} = \binom{6}{r} x^{12 - 2r - r} = \binom{6}{r} x^{12 - 3r} \]For term independent of \( x \), power of \( x = 0 \):
\[ 12 - 3r = 0 \implies r = 4 \] \[ T_5 = \binom{6}{4} (x^2)^2 \left(\frac{1}{x}\right)^4 = \binom{6}{4} x^4 x^{-4} = \binom{6}{4} = 15 \]Answer: 15
Solution:
\[ T_{r+1} = \binom{5}{r} (1)^{5-r} (2x)^r = \binom{5}{r} 2^r x^r \]We want coefficient of \( x^3 \), so \( r=3 \):
\[ \binom{5}{3} 2^3 = 10 \times 8 = 80 \]Answer: 80
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