a) The number of terms in the expansion of \( (2x + 5)^7 \) is 8. In general, for an expansion of \( (a + b)^n \), the number of terms is \( n + 1 \).
b) To find the coefficient of the term in \( x^2 \) in the expansion of \( (2x + 5)^7 \), we use the binomial theorem:
\[ \binom{7}{k} (2x)^{7-k} (5)^k \]
We need \( (2x)^{7-k} \) to have \( x^2 \), so \( 7 - k = 2 \), which gives \( k = 5 \).
Now we calculate the coefficient of this term:
\[ \binom{7}{5} (2x)^{2} (5)^5 \]
Simplifying each part:
\[ \binom{7}{5} = \binom{7}{2} = \frac{7!}{2!5!} = \frac{7 \times 6}{2 \times 1} = 21 \]
\[ (2x)^2 = 4x^2 \]
\[ 5^5 = 3125 \]
Multiplying these together gives the coefficient:
\[ 21 \times 4 \times 3125 = 21 \times 12500 = 262500 \]
Therefore, the coefficient of the term in \( x^2 \) is 262500.
CloseFor the third term the equation is:
\[ \binom{7}{4} x^4 a^{7-4} = 280x^4 \]
Calculate \( \binom{7}{4} \):
\[ \binom{7}{4} = 35 \]
So, the equation simplifies to:
\[ 35 \cdot a^3 = 280 \]
Solve for \( a^3 \):
\[ a^3 = \frac{280}{35} = 8 \]
Therefore, the value of \( a \) is 2.
CloseThe expanded form of \( (2x + 3)^4 \) is:
\[ (2x + 3)^4 = 16x^4 + 4 * 24x^3 + 6 * 36x^2 + 4 * 54x + 81 = 16x^4 + 96x^3 + 216x^2 + 216x + 81 \]
We start with the expansion \( \frac{(x + p)^6}{qx^2} \). The term involving \( x^3 \) can be found as follows:
To find the coefficient of \( x^3 \), we need the term in \( (x + p)^6 \) that becomes \( x^5 \) after dividing by \( x^2 \).
The general term in the expansion of \( (x + p)^6 \) is \( \binom{6}{k} x^{6-k} p^k \).
So, we need the term where \( 6 - k = 5 \), thus \( k = 1 \).
So, the term in \( (x + p)^6 \) is \( \binom{6}{1} x^5 p^1 = 6x^5 p \).
After dividing by \( x^2 \), we get \( \frac{6x^5 p}{qx^2} = \frac{6p}{q} x^3 \).
We know the coefficient of \( x^3 \) is 1, so:
\[ \frac{6p}{q} = 1 \Rightarrow 6p = q \]
Next, for the term involving \( x^2 \), we need the term in \( (x + p)^6 \) that becomes \( x^4 \) after dividing by \( x^2 \).
We need the term where \( 6 - k = 4 \), thus \( k = 2 \).
The term in \( (x + p)^6 \) is \( \binom{6}{2} x^4 p^2 = 15x^4 p^2 \).
After dividing it by \( x^2 \), we get \( \frac{15x^4 p^2}{qx^2} = \frac{15p^2}{q} x^2 \).
We know the coefficient of \( x^2 \) is 5, so:
\[ \frac{15p^2}{q} = 5 \Rightarrow 15p^2 = 5q \Rightarrow 3p^2 = q \]
Substituting \( q = 6p \) into \( 3p^2 = q \), we get:
\[ 3p^2 = 6p \Rightarrow p^2 = 2p \Rightarrow p(p - 2) = 0 \]
Since \( p > 0 \), which is indicated by the question, we have that \( p = 2 \).
Then, \( q = 6p \implies 6 * 2 = 12 \).
Thus, the values are \( p = 2 \) and \( q = 12 \).
a) To find the number of terms in the expansion of \( (x^4 + \frac{4}{x})^9 \), we recognize that each term is of the form \( \binom{9}{k} (x^4)^{9-k} \left(\frac{4}{x}\right)^k \). The total number of terms is given by \( 9 + 1 = 10 \).
b) To find the coefficient of \( x^{16} \), we use the general term of the binomial expansion:
\[ \binom{9}{k} (x^4)^{9-k} \left(\frac{4}{x}\right)^k = \binom{9}{k} (x)^{36-4k} \left(\frac{4^k}{x^k}\right) = \binom{9}{k} * 4^k * x^{36-5k}\]
We need the exponent of \( x \) to be 16:
\[ 36 - 5k = 16 \]
Solving for \( k \):
\[ 36 - 16 = 5k \]
\[ 20 = 5k \]
\[ k = \frac{20}{5} = 4 \]
Now, calculate the coefficient:
\[ \text{Coefficient of } x^{16} = \binom{9}{4} * 4^4 \]
\[ = 126 * 256 \]
\[ = 32256 \]
CloseUsing \( a + b = 5t \):
\[ 15t^2 + 20t^3 = 5t \]
\[ 3t + 4t^2 = 1 \]
\[ (4t-1)(t+1) = 0 \]
Therefore, the possible values of \( t \) are \( -1,\frac{1}{4} \).
CloseWith the binomial theorem we get that:
\[ ax^3(\binom{5}{3} *4^2 * (ax)^3) = 3840x^6 \]
\[ ax^3(\binom{5}{3} *4^2 * (ax)^3) = 3840x^6 \]
\[ ax^3(15 * 16 * a^3x^3) = 3840x^6 \]
\[ 240a^4x^6 = 3840x^6 \]
\[ 240a^4 = 3840 \]
\[ a^4 = 16 \]
So, since \( a < 0 \), then \( a = -2 \).
CloseGiven the coefficient of \( x^4 \) in the expansion of \( (px^2 + q)^6 \) is 30:
We need the term where \( \binom{6}{4} (px^2)^2 q^{4} = \binom{6}{4} p^2 x^4 q^4 \).
\(\binom{6}{4} = 15\), so \( 15 p^2 q^4 = 30 \). Hence, \( p^2 q^4 = 2 \).
The, given the coefficient of \( x^4 \) in the expansion of \( (px^2 + q)^8 \) is 112:
The term we will need is \( \binom{8}{6} (px^2)^2 q^{6} \)
\(\binom{8}{2} = 28\), so \( 28 p^2 q^6 = 112 \). Hence, \( p^2 q^6 = 4 \).
Solving the system of equations:
\[ (1) \ p^2 q^4 = 2 \]
\[ (2) \ p^2 q^6 = 4 \]
\[ p^2 q^4 = 2 \implies p^2 = \frac{2}{q^4} \]
By plugging this into the second equation:
\[ \frac{2}{q^4} * q^6 = 4\]
\[ 2q^2 = 4\]
Knowing that \( q > 0 \):
\[ q = \sqrt{2}\]
\[ p^2 = \frac{1}{2} \implies p = \frac{\sqrt{2}}{2}\]
Therefore, the values are \( p = \frac{\sqrt{2}}{2} \) and \( q = \sqrt{2} \).
CloseWe consider the term \(\binom{5}{3} x^3 k^2\).
Since \(\binom{5}{3} = 10\), we have:
\[ 10x^3 k^2 = 40x^3 \]
Dividing both sides by \(10x^3\), we get:
\[ k^2 = 4 \]
Taking the square root:
\[ k = \pm 2 \]
CloseAt \( x^2 \), the expansion of \( (2x+1)^k \) is:
\[ \binom{k}{2} ( (2x)^2 ) [1]^{(k-2)} \]
Thus, we have:
\[ 20kx^3 = x \left[ \frac{k!}{2!(k-2)!} \right] [4x^2] \]
Rewriting:
\[ 20kx^3 = \left[ \frac{k(k-1)}{2} \right] [4x^3] \]
Solving for \( k \):
\[ 10k = k(k-1) \]
Thus, \( k = 11 \).
CloseThe 5th power of \( x \) will be given from:
\[ kx^2 \times \binom{8}{3} (5^5) (kx)^3 \]
Simplifying:
\[ 22680x^5 = 56 \times 3125 \times k^4 x^5 \]
Solving for \( k^4 \):
\[ k^4 = \frac{22680}{56 \times 3125} \]
\[ k = \frac{3}{5} \]
Closea) The general term in the expansion of \( \left( 2x^3 - \frac{3}{x^2} \right)^{10} \) is given by:
\[ T_{k+1} = \binom{10}{k} \left(2x^3\right)^{10-k} \left(-\frac{3}{x^2}\right)^k \]
Simplify the exponents of \( x \):
\[ T_{k+1} = \binom{10}{k} \cdot 2^{10-k} \cdot (-3)^k \cdot x^{30 - 5k} \]
To find the constant term, set the exponent of \( x \) to zero:
\[ 30 - 5k = 0 \implies k = 6 \]
Substitute \( k = 6 \) into the general term:
\[ T_{7} = \binom{10}{6} \cdot 2^{4} \cdot (-3)^6 \cdot x^{0} = 210 \cdot 16 \cdot 729 = 2449440 \]
Thus, the constant term is \( 2449440 \).
b) Set the exponent of \( x \) to 6:
\[ 30 - 5k = 6 \implies 5k = 24 \implies k = \frac{24}{5} \]
Since \( k \) must be an integer, there is no term with \( x^6 \) in the expansion. Therefore, the coefficient of \( x^6 \) is \( 0 \).
The first three terms correspond to \( k = 0 \), \( k = 1 \), and \( k = 2 \).
For \( k = 0 \):
\[ T_{1} = \binom{10}{0} \cdot 2^{10} \cdot (-3)^0 \cdot x^{30} = 1 \cdot 1024 \cdot 1 \cdot x^{30} = 1024x^{30} \]
For \( k = 1 \):
\[ T_{2} = \binom{10}{1} \cdot 2^{9} \cdot (-3)^1 \cdot x^{25} = 10 \cdot 512 \cdot (-3) \cdot x^{25} = -15360x^{25} \]
For \( k = 2 \):
\[ T_{3} = \binom{10}{2} \cdot 2^{8} \cdot (-3)^2 \cdot x^{20} = 45 \cdot 256 \cdot 9 \cdot x^{20} = 103680x^{20} \]
Thus, the first three terms in descending powers of \( x \) are:
\[ 1024x^{30}, \quad -15360x^{25}, \quad 103680x^{20} \]
CloseTo find the term in \( x^3 \) in the expansion of \( (1 + 2x)^5 \left( 4 + \frac{3}{x} \right) \), we will expand each part and then multiply them together.
Using the binomial theorem:
\[ (1 + 2x)^5 = \sum_{k=0}^5 \binom{5}{k} (1)^{5-k} (2x)^k = \sum_{k=0}^5 \binom{5}{k} 2^k x^k \]
The terms in the expansion are:
\[ \binom{5}{0} 2^0 x^0 = 1 \]
\[ \binom{5}{1} 2^1 x^1 = 10x \]
\[ \binom{5}{2} 2^2 x^2 = 40x^2 \]
\[ \binom{5}{3} 2^3 x^3 = 80x^3 \]
\[ \binom{5}{4} 2^4 x^4 = 80x^4 \]
\[ \binom{5}{5} 2^5 x^5 = 32x^5 \]
Thus:
\[ (1 + 2x)^5 = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5 \]
Multiply each term in the expansion of \( (1 + 2x)^5 \) by \( 4 \) and \( \frac{3}{x} \):
\[ (1 + 2x)^5 \left( 4 + \frac{3}{x} \right) = (1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5) \left( 4 + \frac{3}{x} \right) \]
We need to find the combinations of terms from \( (1 + 2x)^5 \) and \( \left( 4 + \frac{3}{x} \right) \) that result in \( x^3 \).
The contributing terms are:
Thus, the total coefficient of \( x^3 \) is:
\[ 320 + 240 = 560 \]
The term in \( x^3 \) in the expansion of \( (1 + 2x)^5 \left( 4 + \frac{3}{x} \right) \) is \( 560x^3 \).
Closea) To factorize \( 3x^2 + 4x - 4 \), we look for two numbers that multiply to \( 3 \times (-4) = -12 \) and add to \( 4 \). These numbers are \( 6 \) and \( -2 \).
Rewrite the middle term using these numbers:
\[ 3x^2 + 6x - 2x - 4 \]
Factor by grouping:
\[ 3x(x + 2) - 2(x + 2) = (3x - 2)(x + 2) \]
Thus, the quadratic expression \( 3x^2 + 4x - 4 \) can be written as:
\[ 3x^2 + 4x - 4 = (3x - 2)(x + 2) \]
b) Using the factorization from part (a):
\[ (3x^2 + 4x - 4)^6 = [(3x - 2)(x + 2)]^6 = (3x - 2)^6 (x + 2)^6 \]
We need to find the coefficient of \( x \) in the expansion of \( (3x - 2)^6 (x + 2)^6 \).
Using the binomial theorem:
\[ (3x - 2)^6 = \sum_{k=0}^6 \binom{6}{k} (3x)^k (-2)^{6-k} \]
\[ (x + 2)^6 = \sum_{m=0}^6 \binom{6}{m} x^m 2^{6-m} \]
We need to find the combinations of terms from \( (3x - 2)^6 \) and \( (x + 2)^6 \) that result in \( x \).
The exponent of \( x \) in the product \( (3x)^k x^m \) is \( k + m \). We need \( k + m = 1 \).
The possible pairs \((k, m)\) are:
1. For \( k = 0 \), \( m = 1 \):
\[ \binom{6}{0} (3x)^0 (-2)^6 \cdot \binom{6}{1} x^1 2^{5} = 1 \cdot 1 \cdot 64 \cdot 6 \cdot x \cdot 32 = 64 \cdot 6 \cdot 32 \cdot x = 12288x \]
2. For \( k = 1 \), \( m = 0 \):
\[ \binom{6}{1} (3x)^1 (-2)^5 \cdot \binom{6}{0} x^0 2^{6} = 6 \cdot 3x \cdot (-32) \cdot 1 \cdot 1 \cdot 64 = 6 \cdot 3 \cdot (-32) \cdot 64 \cdot x = -36864x \]
The total coefficient of \( x \) is:
\[ 12288 - 36864 = -24576 \]
The coefficient of \( x \) in the expansion of \( (3x^2 + 4x - 4)^6 \) is \( -24576 \).
Close