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 \( q = \sqrt{2} \).
Close