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The Binomial Theorem provides a powerful way to expand expressions of the form \( (a + b)^n \) where \( n \) is a non-negative integer. Instead of multiplying the binomial repeatedly, the theorem gives a direct formula for the expansion.
The expansion of \( (a + b)^n \) is given by:
\[(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\]Here, \( \binom{n}{r} \) is the binomial coefficient, representing the number of ways to choose \( r \) elements from \( n \), and is calculated as:
\[\binom{n}{r} = \frac{n!}{r! (n-r)!}\]Binomial coefficients correspond to the entries in Pascal's Triangle. Each row \( n \) of Pascal's Triangle gives the coefficients for \( (a + b)^n \).
For example, the expansion of \( (a + b)^4 \) is:
\[(a + b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4\]The \((r+1)^\text{th}\) term in the expansion of \( (a + b)^n \) is:
\[T_{r+1} = \binom{n}{r} a^{n-r} b^r\]This formula is useful for finding any specific term without expanding the entire expression.
The binomial theorem is widely used in algebra, probability, combinatorics, and even calculus for approximations.
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
Binomial coefficients also represent the number of subsets of size \( r \) from a set of size \( n \). This is directly linked to the concept of power sets and subsets in set theory.
For example, if a set \( S \) has \( n \) elements, then the number of subsets with exactly \( r \) elements is \( \binom{n}{r} \). The total number of subsets (the power set) is \( 2^n \).
Expand \( (x + 2)^3 \) using the Binomial Theorem.
Solution:
Here, \( n = 3 \), \( a = x \), \( b = 2 \).
\[ (x + 2)^3 = \sum_{r=0}^3 \binom{3}{r} x^{3-r} (2)^r \]Calculate each term:
\[ \binom{3}{0} x^3 2^0 = 1 \cdot x^3 \cdot 1 = x^3 \]Final expansion:
\[ (x + 2)^3 = x^3 + 6x^2 + 12x + 8 \]Find the 4th term of the expansion of \( (2x - 3)^5 \).
Solution:
General term \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \).
Here, \( n=5 \), \( a=2x \), \( b = -3 \), and we want the 4th term, so \( r=3 \).
\[ T_4 = \binom{5}{3} (2x)^{5-3} (-3)^3 = \binom{5}{3} (2x)^2 (-27) \] \[ = 10 \times 4x^2 \times (-27) = 10 \times 4 \times (-27) x^2 = -1080 x^2 \]Answer: \( -1080 x^2 \)
Find the sum of the coefficients in the expansion of \( (3x + 2)^4 \).
Solution:
Sum of coefficients is found by putting \( x=1 \):
\[ (3 \times 1 + 2)^4 = (3 + 2)^4 = 5^4 = 625 \]Answer: 625
Find the term independent of \( x \) in the expansion of \( \left( x^2 + \frac{1}{x} \right)^6 \).
Solution:
General term:
\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} \left(\frac{1}{x}\right)^r = \binom{6}{r} x^{2(6-r)} x^{-r} = \binom{6}{r} x^{12 - 2r - r} = \binom{6}{r} x^{12 - 3r} \]Term independent of \( x \) means power of \( x = 0 \):
\[ 12 - 3r = 0 \implies 3r = 12 \implies r = 4 \]Calculate the term for \( r=4 \):
\[ T_5 = \binom{6}{4} (x^2)^2 \left(\frac{1}{x}\right)^4 = \binom{6}{4} x^{4} x^{-4} = \binom{6}{4} = 15 \]Answer: 15
Find the coefficient of \( x^3 \) in the expansion of \( (1 + 2x)^5 \).
Solution:
General term:
\[ T_{r+1} = \binom{5}{r} 1^{5-r} (2x)^r = \binom{5}{r} 2^r x^r \]We want the term where power of \( x \) is 3, so \( r=3 \):
\[ \binom{5}{3} 2^3 x^3 = 10 \times 8 \times x^3 = 80 x^3 \]Coefficient of \( x^3 \) is 80.
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