Definition: Binomial
A binomial is an algebraic expression that contains exactly two terms connected by a plus or minus sign.
Examples: \( (a + b), \quad (x - y), \quad (3x + 2) \)
Applications of Binomials:
A binomial is an algebraic expression that contains exactly two terms connected by a plus or minus sign.
Examples: \( (a + b), \quad (x - y), \quad (3x + 2) \)
Applications of Binomials:
- Used extensively in algebra to simplify and expand expressions.
- Foundation for the Binomial Theorem, which helps in expanding powers of binomials.
- Applied in probability theory, especially in the binomial distribution.
- Useful in calculus, particularly in series expansions and approximations.
General Formula: Binomial Expansion
For any positive integer \( n \), the expansion of the binomial \( (a + b)^n \) is given by:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \] \[ = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \cdots + \binom{n}{n} a^0 b^n \]
where \(\displaystyle \binom{n}{k}\) is the binomial coefficient:
\[ \binom{n}{k} = \frac{n!}{k! (n-k)!} \]
This formula allows you to expand any binomial raised to a positive integer power without multiplying repeatedly.
For any positive integer \( n \), the expansion of the binomial \( (a + b)^n \) is given by:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \] \[ = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \cdots + \binom{n}{n} a^0 b^n \]
where \(\displaystyle \binom{n}{k}\) is the binomial coefficient:
\[ \binom{n}{k} = \frac{n!}{k! (n-k)!} \]
This formula allows you to expand any binomial raised to a positive integer power without multiplying repeatedly.
Example 1: Binomial Expansion
Expand \( (x + 2)^3 \):
Using the binomial theorem:
Calculating terms:
Expand \( (x + 2)^3 \):
Using the binomial theorem:
\[
(x + 2)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} (2)^k = \binom{3}{0} x^3 (2)^0 + \binom{3}{1} x^2 (2)^1 + \binom{3}{2} x^1 (2)^2 + \binom{3}{3} x^0 (2)^3
\]
Calculating terms:
\[
= 1 \cdot x^3 + 3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 4 + 1 \cdot 8 = x^3 + 6x^2 + 12x + 8
\]
Example 2:
To find the 5th term in the expansion of \( \displaystyle \left( x + \frac{1}{x} \right)^5 \), we use the Binomial Theorem:
General term:
\( T_{r+1} = \binom{5}{r} \cdot x^{5 - 2r} \)
For the 5th term, let \( r = 4 \):
\( T_5 = \binom{5}{4} \cdot x^{5 - 2 \times 4} = 5x^{-3} \)
Final Answer: \( \boxed{5x^{-3}} \)
To find the 5th term in the expansion of \( \displaystyle \left( x + \frac{1}{x} \right)^5 \), we use the Binomial Theorem:
General term:
\( T_{r+1} = \binom{5}{r} \cdot x^{5 - 2r} \)
For the 5th term, let \( r = 4 \):
\( T_5 = \binom{5}{4} \cdot x^{5 - 2 \times 4} = 5x^{-3} \)
Final Answer: \( \boxed{5x^{-3}} \)
Example 3: Expand \( (2x - 3)^4 \)
Using the binomial theorem:
Calculating terms:
Using the binomial theorem:
\[
(2x - 3)^4 = \sum_{k=0}^4 \binom{4}{k} (2x)^{4-k} (-3)^k
\]
Calculating terms:
\[
= \binom{4}{0} (2x)^4 (-3)^0 + \binom{4}{1} (2x)^3 (-3)^1 + \binom{4}{2} (2x)^2 (-3)^2 + \binom{4}{3} (2x)^1 (-3)^3 + \binom{4}{4} (2x)^0 (-3)^4
\]
\[
= 1 \cdot 16x^4 \cdot 1 - 4 \cdot 8x^3 \cdot 3 + 6 \cdot 4x^2 \cdot 9 - 4 \cdot 2x \cdot 27 + 1 \cdot 1 \cdot 81
\]
\[
= 16x^4 - 96x^3 + 216x^2 - 216x + 81
\]
Example 4: Find the coefficient of \( x^2 \) in \( (1 + 3x)^5 \)
The general term:
Coefficient of \( x^2 \) set \( k=2 \)
The general term:
\[
T_{k+1} = \binom{5}{k} (1)^{5-k} (3x)^k = \binom{5}{k} 3^k x^k
\]
Coefficient of \( x^2 \) set \( k=2 \)
\[
\binom{5}{2} 3^2 = 10 \times 9 = 90
\]
Example 5: Find the term independent of \( x \) in \( \left(x + \frac{2}{x}\right)^5 \)
General term:
Term independent of \( x \) means power of \( x \) is zero:
Conclusion: No term independent of \( x \) exists in this expansion.
General term:
\[
T_{k+1} = \binom{5}{k} x^{5-k} \left(\frac{2}{x}\right)^k = \binom{5}{k} 2^k x^{5 - 2k}
\]
Term independent of \( x \) means power of \( x \) is zero:
\[
5 - 2k = 0 \implies k = \frac{5}{2} \quad \text{(not an integer, so no independent term)}
\]
Conclusion: No term independent of \( x \) exists in this expansion.
Example 6: Find the term independent of \( x \) in \( \left(2x^2 + \frac{3}{x}\right)^6 \)
General term:
Term independent of \( x \) means power of \( x \) is zero:
Substitute \( k = 4 \):
Answer: Term independent of \( x \) is \( \boxed{4860} \).
General term:
\[
T_{k+1} = \binom{6}{k} (2x^2)^{6-k} \left(\frac{3}{x}\right)^k = \binom{6}{k} 2^{6-k} 3^{k} x^{12 - 3k}
\]
Term independent of \( x \) means power of \( x \) is zero:
\[
12 - 3k = 0 \implies k = 4
\]
Substitute \( k = 4 \):
\[
T_5 = \binom{6}{4} 2^{2} 3^{4} = 15 \times 4 \times 81 = 4860
\]
Answer: Term independent of \( x \) is \( \boxed{4860} \).
Example 7: Find the term independent of \( x \) in \( \left( \frac{1}{\sqrt{x}} + x^2 \right)^5 \)
General term:
\[ T_{k+1} = \binom{5}{k} \left( \frac{1}{\sqrt{x}} \right)^{5-k} (x^2)^k = \binom{5}{k} x^{-\frac{5-k}{2}} \cdot x^{2k} \]
Combine the powers of \(x\):
\[ x^{-\frac{5-k}{2} + 2k} = x^{\left( -\frac{5-k}{2} + 2k \right)} = x^{\left( -\frac{5}{2} + \frac{k}{2} + 2k \right)} = x^{\left( \frac{5k - 5}{2} \right)} \]
To be independent of \(x\), exponent must be 0:
\[ \frac{5k - 5}{2} = 0 \Rightarrow 5k - 5 = 0 \Rightarrow k = 1 \]
So, the required term is:
\[ T_2 = \binom{5}{1} \left( \frac{1}{\sqrt{x}} \right)^4 (x^2)^1 = 5 \cdot x^{-2} \cdot x^2 = 5 \]
Final Answer: \( \boxed{5} \)
General term:
\[ T_{k+1} = \binom{5}{k} \left( \frac{1}{\sqrt{x}} \right)^{5-k} (x^2)^k = \binom{5}{k} x^{-\frac{5-k}{2}} \cdot x^{2k} \]
Combine the powers of \(x\):
\[ x^{-\frac{5-k}{2} + 2k} = x^{\left( -\frac{5-k}{2} + 2k \right)} = x^{\left( -\frac{5}{2} + \frac{k}{2} + 2k \right)} = x^{\left( \frac{5k - 5}{2} \right)} \]
To be independent of \(x\), exponent must be 0:
\[ \frac{5k - 5}{2} = 0 \Rightarrow 5k - 5 = 0 \Rightarrow k = 1 \]
So, the required term is:
\[ T_2 = \binom{5}{1} \left( \frac{1}{\sqrt{x}} \right)^4 (x^2)^1 = 5 \cdot x^{-2} \cdot x^2 = 5 \]
Final Answer: \( \boxed{5} \)
Example 8: Find the term containing \( x^3 y^2 \) in the expansion of \( (2x - 5y)^5 \)
The general term in \( (a + b)^n \) is:
Powers of \(x\) and \(y\) are \(x^{5-k}\) and \(y^k\).
To get \(x^3 y^2\), solve \(5-k=3 \Rightarrow k=2\).
Substitute \(k=2\):
Calculate coefficients:
\[ \binom{5}{2} = 10, \quad 2^3 = 8, \quad (-5)^2 = 25 \] Multiply:
\[ 10 \times 8 \times 25 = 2000 \] So the term is:
\[ 2000 x^3 y^2 \] Final answer: \( \boxed{2000 x^3 y^2} \)
The general term in \( (a + b)^n \) is:
\[
T_{k+1} = \binom{5}{k} (2x)^{5-k} (-5y)^k
\]
Powers of \(x\) and \(y\) are \(x^{5-k}\) and \(y^k\).
To get \(x^3 y^2\), solve \(5-k=3 \Rightarrow k=2\).
Substitute \(k=2\):
\[
T_3 = \binom{5}{2} (2x)^3 (-5y)^2
\]
Calculate coefficients:
\[ \binom{5}{2} = 10, \quad 2^3 = 8, \quad (-5)^2 = 25 \] Multiply:
\[ 10 \times 8 \times 25 = 2000 \] So the term is:
\[ 2000 x^3 y^2 \] Final answer: \( \boxed{2000 x^3 y^2} \)
Example 9: Find the term containing \( x^2 y^3 \) in the expansion of \( (3x + 4y)^5 \)
The general term in \( (a + b)^n \) is:
Powers of \(x\) and \(y\) are \(x^{5-k}\) and \(y^k\).
To get \(x^2 y^3\), solve \(5-k=2 \Rightarrow k=3\).
Substitute \(k=3\):
Calculate coefficients:
\[ \binom{5}{3} = 10, \quad 3^2 = 9, \quad 4^3 = 64 \] Multiply:
\[ 10 \times 9 \times 64 = 5760 \] So the term is:
\[ 5760 x^2 y^3 \] Final answer: \( \boxed{5760 x^2 y^3} \)
The general term in \( (a + b)^n \) is:
\[
T_{k+1} = \binom{5}{k} (3x)^{5-k} (4y)^k
\]
Powers of \(x\) and \(y\) are \(x^{5-k}\) and \(y^k\).
To get \(x^2 y^3\), solve \(5-k=2 \Rightarrow k=3\).
Substitute \(k=3\):
\[
T_4 = \binom{5}{3} (3x)^2 (4y)^3
\]
Calculate coefficients:
\[ \binom{5}{3} = 10, \quad 3^2 = 9, \quad 4^3 = 64 \] Multiply:
\[ 10 \times 9 \times 64 = 5760 \] So the term is:
\[ 5760 x^2 y^3 \] Final answer: \( \boxed{5760 x^2 y^3} \)
Converting Term Number from Last to Term Number from Start in Binomial Expansion
In the expansion of \( (a + b)^n \), there are \( n+1 \) terms.
Terms are numbered from the start as \( T_1, T_2, \ldots, T_{n+1} \).
The \( r^\text{th} \) term from the last corresponds to the term number \( k \) from the start, given by:
\[ k = (n + 2) - r \] where
\( n \) = power of the binomial,
\( r \) = term number from the last,
\( k \) = term number from the start.
Example 10: Find the 3rd term from the last in \( (2x - 5y)^5 \).
Here, \( n = 5 \) and \( r = 3 \).
Using the formula:
\[ k = (5 + 2) - 3 = 4 \]
So, the 3rd term from the last is the 4th term from the start.
This conversion is useful because the binomial theorem formula for the general term \( T_{k} \) is counted from the start.
In the expansion of \( (a + b)^n \), there are \( n+1 \) terms.
Terms are numbered from the start as \( T_1, T_2, \ldots, T_{n+1} \).
The \( r^\text{th} \) term from the last corresponds to the term number \( k \) from the start, given by:
\[ k = (n + 2) - r \] where
\( n \) = power of the binomial,
\( r \) = term number from the last,
\( k \) = term number from the start.
Example 10: Find the 3rd term from the last in \( (2x - 5y)^5 \).
Here, \( n = 5 \) and \( r = 3 \).
Using the formula:
\[ k = (5 + 2) - 3 = 4 \]
So, the 3rd term from the last is the 4th term from the start.
This conversion is useful because the binomial theorem formula for the general term \( T_{k} \) is counted from the start.
Example 11: Find the 3rd term from the last in the expansion of \( (2x - 5y)^5 \)
Total terms = 6 (since \(n=5\))
The 3rd term from the last is the \((6 - 3 + 1) = 4^\text{th}\) term from the start.
General term:
For the 4th term, \(r = 3\):
Calculate coefficients:
\[ \binom{5}{3} = 10, \quad 2^2 = 4, \quad (-5)^3 = -125 \] Multiply:
\[ 10 \times 4 \times (-125) = -5000 \] So the term is:
\[ -5000 x^{2} y^{3} \] Final answer: \( \boxed{-5000 x^{2} y^{3}} \)
Total terms = 6 (since \(n=5\))
The 3rd term from the last is the \((6 - 3 + 1) = 4^\text{th}\) term from the start.
General term:
\[
T_{r+1} = \binom{5}{r} (2x)^{5-r} (-5y)^r
\]
For the 4th term, \(r = 3\):
\[
T_4 = \binom{5}{3} (2x)^{2} (-5y)^3
\]
Calculate coefficients:
\[ \binom{5}{3} = 10, \quad 2^2 = 4, \quad (-5)^3 = -125 \] Multiply:
\[ 10 \times 4 \times (-125) = -5000 \] So the term is:
\[ -5000 x^{2} y^{3} \] Final answer: \( \boxed{-5000 x^{2} y^{3}} \)
Example 12: Find the 2nd term from the last in the expansion of \( (3x + 4y)^6 \)
Total terms = 7 (since \(n=6\))
The 2nd term from the last is the \((7 - 2 + 1) = 6^\text{th}\) term from the start.
General term:
For the 6th term, \(r = 5\):
Calculate coefficients:
\[ \binom{6}{5} = 6, \quad 3^{1} = 3, \quad 4^{5} = 1024 \] Multiply:
\[ 6 \times 3 \times 1024 = 18432 \] So the term is:
\[ 18432 x^{1} y^{5} \] Final answer: \( \boxed{18432 x y^{5}} \)
Total terms = 7 (since \(n=6\))
The 2nd term from the last is the \((7 - 2 + 1) = 6^\text{th}\) term from the start.
General term:
\[
T_{r+1} = \binom{6}{r} (3x)^{6-r} (4y)^r
\]
For the 6th term, \(r = 5\):
\[
T_6 = \binom{6}{5} (3x)^{1} (4y)^5
\]
Calculate coefficients:
\[ \binom{6}{5} = 6, \quad 3^{1} = 3, \quad 4^{5} = 1024 \] Multiply:
\[ 6 \times 3 \times 1024 = 18432 \] So the term is:
\[ 18432 x^{1} y^{5} \] Final answer: \( \boxed{18432 x y^{5}} \)
How to Find the Middle Term(s) in a Binomial Expansion
Consider the expansion of \( (a + b)^n \). The total number of terms is \( n + 1 \).
Consider the expansion of \( (a + b)^n \). The total number of terms is \( n + 1 \).
- If \(n + 1\) is odd:
There is exactly one middle term at position \( \frac{n + 2}{2} \). - If \(n + 1\) is even:
There are two middle terms at positions \( \frac{n + 1}{2} \) and \( \frac{n + 1}{2} + 1 \).
Example 13: Middle Terms of \( (x + 2)^5 \) (Even Number of Terms)
Total terms \(= 5 + 1 = 6\) (even)
Middle terms are at positions:
\[ \frac{6}{2} = 3, \quad \frac{6}{2} + 1 = 4 \]
So, the 3rd and 4th terms are the middle terms.
General term formula:
\[ T_{k+1} = \binom{5}{k} x^{5-k} (2)^k \]
Calculate terms:
\[ T_3 = \binom{5}{2} x^{3} (2)^2 = 10 \times x^{3} \times 4 = 40 x^{3} \]
\[ T_4 = \binom{5}{3} x^{2} (2)^3 = 10 \times x^{2} \times 8 = 80 x^{2} \]
Middle terms: \(40 x^{3}\) and \(80 x^{2}\).
Total terms \(= 5 + 1 = 6\) (even)
Middle terms are at positions:
\[ \frac{6}{2} = 3, \quad \frac{6}{2} + 1 = 4 \]
So, the 3rd and 4th terms are the middle terms.
General term formula:
\[ T_{k+1} = \binom{5}{k} x^{5-k} (2)^k \]
Calculate terms:
\[ T_3 = \binom{5}{2} x^{3} (2)^2 = 10 \times x^{3} \times 4 = 40 x^{3} \]
\[ T_4 = \binom{5}{3} x^{2} (2)^3 = 10 \times x^{2} \times 8 = 80 x^{2} \]
Middle terms: \(40 x^{3}\) and \(80 x^{2}\).
Example 14: Middle Term of \( (3x - y)^4 \) (Odd Number of Terms)
Total terms \(= 4 + 1 = 5\) (odd)
Middle term is at position:
\[ \frac{5 + 1}{2} = 3 \]
General term formula:
\[ T_{k+1} = \binom{4}{k} (3x)^{4-k} (-y)^k \]
Calculate the 3rd term (\(k=2\)):
\[ T_3 = \binom{4}{2} (3x)^2 (-y)^2 = 6 \times 9 x^2 \times y^2 = 54 x^{2} y^{2} \]
Middle term: \(54 x^{2} y^{2}\).
Total terms \(= 4 + 1 = 5\) (odd)
Middle term is at position:
\[ \frac{5 + 1}{2} = 3 \]
General term formula:
\[ T_{k+1} = \binom{4}{k} (3x)^{4-k} (-y)^k \]
Calculate the 3rd term (\(k=2\)):
\[ T_3 = \binom{4}{2} (3x)^2 (-y)^2 = 6 \times 9 x^2 \times y^2 = 54 x^{2} y^{2} \]
Middle term: \(54 x^{2} y^{2}\).
Binomial Expansion: 10 Examples Each
Term Containing \( x^m y^n \)
1. \( (2x - 5y)^5 \), term containing \( x^3 y^2 \): \(2000\)
2. \( (3x + 4y)^5 \), term containing \( x^2 y^3 \): \(5760\)
3. \( (x + y)^6 \), term containing \( x^4 y^2 \): \(15\)
4. \( (2x + 3y)^4 \), term containing \( x^1 y^3 \): \(108\)
5. \( (x - 2y)^7 \), term containing \( x^5 y^2 \): \(-3360\)
6. \( (4x + y)^3 \), term containing \( x^2 y^1 \): \(48\)
7. \( (x + 2y)^8 \), term containing \( x^6 y^2 \): \(8960\)
8. \( (3x - y)^5 \), term containing \( x^3 y^2 \): \(-540\)\)
9. \( (x + 4y)^7 \), term containing \( x^4 y^3 \): \(13440\)
10. \( (2x - 3y)^6 \), term containing \( x^3 y^3 \): \(-8640\)
Term Independent of \( x \)
1. \( (2x^2 + \frac{3}{x})^6 \): \(4860\)
2. \( \left(\frac{1}{\sqrt{x}} + x^2\right)^5 \): \(80\)
3. \( (x + \frac{2}{x})^5 \): No independent term
4. \( \left(x^3 + \frac{1}{x^2}\right)^7 \): \(35\)
5. \( \left(3x + \frac{2}{x^2}\right)^6 \): \(120\)\)
6. \( \left( \frac{2}{x} + x^4 \right)^5 \): \(80\)
7. \( (x^2 + \frac{1}{x^3})^8 \): \(560\)\)
8. \( \left(\frac{1}{x^3} + 2x^2\right)^7 \): \(140\)\)
9. \( (x + \frac{1}{x})^{10} \): \(252\)
10. \( \left(2x^2 + \frac{1}{x^4}\right)^6 \): \(15\)
Term from Last in Expansion
1. 3rd term from last in \( (2x - 5y)^5 \): \(-5000 x^{2} y^{3}\)
2. 2nd term from last in \( (3x + 4y)^6 \): \(18432 x y^{5}\)
3. 4th term from last in \( (x + y)^7 \): \(210 x^{4} y^{3}\)
4. Last term in \( (2x - 3y)^5 \): \(243 y^{5}\)
5. 5th term from last in \( (x - 2y)^8 \): \(-1792 x^{4} y^{4}\)
6. 2nd term from last in \( (3x + y)^4 \): \(54 x y^{3}\)
7. Last term in \( (x + 5y)^6 \): \(15625 y^{6}\)
8. 3rd term from last in \( (4x - y)^5 \): \(-960 x^{2} y^{3}\)
9. 2nd term from last in \( (x + 2y)^7 \): \(448 x y^{6}\)
10. Last term in \( (5x - 2y)^3 \): \(-8 y^{3}\)
Middle Term(s) in Expansion
1. Middle terms of \( (x + 2)^5 \): \(40 x^{3} \text{ and } 80 x^{2}\)
2. Middle term of \( (3x - y)^4 \): \(54 x^{2} y^{2}\)
3. Middle term of \( (x + y)^7 \): \(35 x^{3} y^{4}\)
4. Middle terms of \( (2x + 3)^6 \): \(240 x^{3} \text{ and } 270 x^{2}\)
5. Middle term of \( (x - 2y)^3 \): \(-12 x^{1} y^{2}\)
6. Middle term of \( (4x + y)^5 \): \(80 x^{2} y^{3}\)
7. Middle term of \( (x + y)^8 \): \(70 x^{4} y^{4}\)
8. Middle terms of \( (3x - 2y)^6 \): \(540 x^{3} y^{3} \text{ and } 270 x^{2} y^{4}\)
9. Middle term of \( (x + 2y)^9 \): \(3360 x^{4} y^{5}\)
10. Middle term of \( (x - y)^{10} \): \(252 x^{5} y^{5}\)
Tricky Examples
1. Term containing \( x^{4} y^{-3} \) in \( (x^{2} + \frac{1}{y^{3}})^5 \): \(10\)
2. Term independent of \( x \) in \( \left(\frac{1}{x^{2}} + x^{3}\right)^6 \): \(540\)
3. Term containing \( x^{3} y^{-1} \) in \( (x^{4} - \frac{1}{y})^5 \): \(-80\)
4. Term independent of \( x \) in \( \left(\frac{2}{x} + x^{4}\right)^5 \): \(80\)
5. Term containing \( x^{6} y^{-5} \) in \( (x^{3} + \frac{1}{y^{5}})^4 \): \(4\)
6. Term independent of \( x \) in \( \left(3x^{2} + \frac{2}{x^{3}}\right)^5 \): \(80\)
7. Term containing \( x^{5} y^{-3} \) in \( (x^{2} + \frac{1}{y^{3}})^5 \): \(0 \text{ (does not exist)}\)
8. Term independent of \( x \) in \( \left(x + \frac{1}{x^{2}}\right)^6 \): \(90\)
9. Term containing \( x^{2} y^{-1} \) in \( (x^{3} + \frac{1}{y})^4 \): \(12\)
10. Term independent of \( x \) in \( (x^{2} + \frac{1}{x^{4}})^7 \): \(105\)
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