Example Question - binomial expansion
Here are examples of questions we've helped users solve.
Calculating Specific Expansion Terms and Investigating Geometric Mean Relation
For the 2nd question (problem 8), since it is cut off, we are unable to provide a complete solution. However, for the 3rd question (problem 9), we can proceed: <p>The sum of the two numbers is $6$ times their geometric mean, then </p> <p>Let the two numbers be $a$ and $b$. Given that $a+b = 6\sqrt{ab}$ and we are to show that $a$ and $b$ are in the ratio $(3 + 2\sqrt{2}):(3 - 2\sqrt{2})$.</p> <p>Since we have a ratio of sums to a geometric mean, let's assume $a = (3 + 2\sqrt{2})k$ and $b = (3 - 2\sqrt{2})k$, where $k$ is some positive constant.</p> <p>\[ a + b = (3 + 2\sqrt{2})k + (3 - 2\sqrt{2})k\]</p> <p>\[ a + b = 3k + 2\sqrt{2}k + 3k - 2\sqrt{2}k\]</p> <p>\[ a + b = 6k\]</p> <p>The geometric mean of $a$ and $b$ is $\sqrt{ab}$:</p> <p>\[\sqrt{ab} = \sqrt{(3 + 2\sqrt{2})k \cdot (3 - 2\sqrt{2})k}\]</p> <p>\[\sqrt{ab} = \sqrt{(9 - 8)k^2}\]</p> <p>\[\sqrt{ab} = k\]</p> <p>Now, we compare the sum and geometric mean:</p> <p>\[6k = 6\sqrt{ab}\]</p> <p>\[k = \sqrt{ab}\]</p> <p>This confirms that the sum of $a$ and $b$ is indeed $6$ times their geometric mean, and thus the numbers are in the desired ratio.</p> For question 9, which discusses mean and standard deviation, no specific mathematical equations are shown, and thus a solution cannot be provided based on the information given in the image.
Question on Sequence Terms from Binomial Expansion and Line Equations
<p>For the binomial expansion part, to find the 4th term from the end in the expansion of $\left( \frac{3}{2} - \frac{x^3}{6} \right)^7$, note that the 4th term from the end is equivalent to the 4th term from the beginning (or the $T_4$ term) because the binomial is symmetric.</p> <p>To find $T_4$, we use the binomial theorem which states $T_{k+1} = ^nC_k \cdot a^{n-k} \cdot b^{k}$.</p> <p>$T_4 = ^7C_3 \left( \frac{3}{2} \right)^{7-3} \left( -\frac{x^3}{6} \right)^3$</p> <p>$T_4 = 35 \cdot \left( \frac{3}{2} \right)^4 \cdot \left( -\frac{x^3}{6} \right)^3$</p> <p>$T_4 = 35 \cdot \frac{81}{16} \cdot -\frac{x^9}{216}$</p> <p>$T_4 = -\frac{35 \cdot 81 \cdot x^9}{16 \cdot 216}$</p> <p>$T_4 = -\frac{35 \cdot 81 \cdot x^9}{6^3 \cdot 6}$</p> <p>$T_4 = -\frac{35 \cdot 81 x^9}{6^4}$</p> <p>$T_4 = -\frac{945 \cdot x^9}{1296}$</p> <p>For the equation of lines passing through (1,2) and making an angle $30^\circ$ with the y-axis, the slope of the line is the tangent of the angle with the x-axis. Since the line makes a $30^\circ$ angle with the y-axis, it makes a $60^\circ$ angle with the x-axis. Hence, the slope $m$ is $\tan(60^\circ) = \sqrt{3}$.</p> <p>The equation of a line in slope-intercept form is $y = mx + b$.</p> <p>To find $b$, substitute $(1,2)$ (x, y) into the equation:</p> <p>$2 = \sqrt{3}(1) + b$</p> <p>$b = 2 - \sqrt{3}$</p> <p>Therefore, the equation of the line is:</p> <p>$y = \sqrt{3}x + (2 - \sqrt{3})$</p>
Analysis of a Mathematical Expansion and Counting Problem
<p>The question you've asked appears to be two separate math problems:</p> <p>1. For the number of words each of 3 vowels and 2 consonants that can be formed from "UTE" - This seems like a permutations problem, but the question seems incomplete as it does not state how many vowels and consonants are available to choose from. Without further information, this part of the question cannot be solved.</p> <p>2. For finding the 3rd term from the end in the expansion of \(\left(\frac{3}{x^2} - x^3\right)^7\) - This can be solved using the binomial theorem which states that \(n\)th term from the end is given by \(T_{(r+1)} = ^nC_r \cdot a^{(n-r)} \cdot b^r\) for the expansion of \((a + b)^n\), where \(n\) is the power and \(r\) is \(n-k\) if you're looking for the \(k\)th term from the end.</p> <p>We're looking for the 3rd term from the end (which is the same as the 5th term from the beginning since there are 7 terms in total), so \(r=7-3=4\). Using this, the term is:</p> <p>\[T_{(4+1)} = ^7C_4 \cdot \left(\frac{3}{x^2}\right)^{(7-4)} \cdot (x^3)^4\]</p> <p>\[T_5 = ^7C_4 \cdot \frac{3^3}{x^6} \cdot x^{12}\]</p> <p>\[T_5 = 35 \cdot \frac{27}{x^6} \cdot x^{12}\]</p> <p>\[T_5 = 35 \cdot 27 \cdot x^6\]</p> <p>\[T_5 = 945x^6\]</p> <p>So the 3rd term from the end in the expansion of \(\left(\frac{3}{x^2} - x^3\right)^7\) is \(945x^6\).</p>
Multiple Mathematics Problems Involving Algebra, Geometry, and Complex Numbers
<p>First question: Find the 4th term from the end in the expansion of \( \left( \frac{3}{x^2} - x^3 \right)^7 \).</p> <p>The 4th term from the end is the same as the 4th term from the beginning, which corresponds to \( T_4 \) in the expansion:</p> <p>\( T_k = \binom{n}{k-1} \cdot (a)^{n-(k-1)} \cdot (b)^{k-1} \) where \( n = 7 \), \( a = \frac{3}{x^2} \), and \( b = -x^3 \).</p> <p>\( T_4 = \binom{7}{3} \cdot \left(\frac{3}{x^2}\right)^{7-3} \cdot (-x^3)^3 \)</p> <p>\( T_4 = 35 \cdot \left(\frac{3}{x^2}\right)^4 \cdot (-x^3)^3 \)</p> <p>\( T_4 = 35 \cdot \frac{81}{x^8} \cdot (-x^9) \)</p> <p>\( T_4 = 35 \cdot 81 \cdot \frac{-x^9}{x^8} \)</p> <p>\( T_4 = -2835 \cdot \frac{1}{x} \)</p> <p>Second question: Find the equation of lines passing through (1,2) and making angle 30° with y-axis.</p> <p>The slope of the line making a 30° angle with the y-axis is the tangent of (90°-30°), which is \( \tan(60°) \).</p> <p>Slope (m) of the desired line: \( m = \tan(60°) = \sqrt{3} \)</p> <p>Use the point-slope form \( y - y_1 = m(x - x_1) \) to find the equation of the line passing through (1,2).</p> <p>\( y - 2 = \sqrt{3}(x - 1) \)</p> <p>Equation of the line: \( y = \sqrt{3}x + (2 - \sqrt{3}) \)</p> <p>Third question: Find the domain and range of the function \( f(x) = \frac{1}{\sqrt{9 - x^2}} \).</p> <p>The domain is where the function is defined and the denominator is not zero.</p> <p>\( 9 - x^2 > 0 \)</p> <p>\( -3 < x < 3 \), so the domain is \( (-3, 3) \).</p> <p>Since the function is the reciprocal of a square root, its range is all positive real numbers, \( (0, \infty) \).</p> <p>Fourth question: Solve the system of equations \( Re(z) = 0, |z| = 2 \).</p> <p>A complex number \( z = x + yi \) satisfies \( Re(z) = 0 \) when \( x = 0 \).</p> <p>Using \( |z| = 2 \), we get \( |0 + yi| = 2 \), which means \( \sqrt{0^2 + y^2} = 2 \).</p> <p>\( y = \pm 2 \), so \( z = 0 \pm 2i \).</p> <p>Fifth question: Write the complex number \( 1 + 7i \) in polar form.</p> <p>Let \( z = 1 + 7i \).</p> <p>Magnitude \( r = |z| = \sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt{2} \).</p> <p>Angle \( \theta = \tan^{-1}\left(\frac{7}{1}\right) \), which is in the first quadrant.</p> <p>Polar form: \( z = r(\cos \theta + i \sin \theta) \)</p> <p>\( z = 5\sqrt{2}\left(\cos \left(\tan^{-1} 7\right) + i \sin \left(\tan^{-1} 7\right)\right) \)</p>
Finding a Specific Term in Binomial Expansion
<p>To find the 4th term from the end in the expansion of \(\left(3x^2 - \frac{x^3}{6}\right)^7\), we can use the General Term formula for binomial expansion:</p> <p>The General Term (T_k) of (a + b)^n is given by:</p> <p>T_k = C(n, k-1) \cdot a^{n-k+1} \cdot b^{k-1}</p> <p>Since we're looking for the 4th term from the end, for n = 7, the term we're looking for is the 7 - 4 + 1 = 4th term (T_4).</p> <p>T_4 = C(7, 4-1) \cdot \left(3x^2\right)^{7-4+1} \cdot \left(-\frac{x^3}{6}\right)^{4-1}</p> <p>T_4 = C(7, 3) \cdot \left(3x^2\right)^4 \cdot \left(-\frac{x^3}{6}\right)^3</p> <p>T_4 = 35 \cdot \left(81x^8\right) \cdot \left(-\frac{x^9}{216}\right)</p> <p>T_4 = 35 \cdot 81 \cdot \left(-\frac{1}{216}\right) \cdot x^{8+9}</p> <p>T_4 = -\frac{35 \cdot 81 \cdot x^{17}}{216}</p> <p>T_4 = -\frac{2835 \cdot x^{17}}{216}</p> <p>T_4 = -\frac{35 \cdot x^{17}}{8}</p>
Simplified Evaluation of Complex Number Expression
The expression you're being asked to evaluate is: (1 + 3i)^8 + (1 - 3i)^8 This is an expression involving complex numbers. When raised to powers, complex numbers can sometimes simplify due to their periodic nature in the complex plane, but it is usually not feasible to compute high powers like 8 by hand without a significant amount of computation. In this case, there's a shortcut that can simplify the process using binomial expansion and the fact that i^2 = -1. To evaluate this expression, you would normally apply the binomial theorem, which states that: (a + b)^n = Σ[k=0 to n] (n choose k) * a^(n-k) * b^k Where "n choose k" is the binomial coefficient, calculated as: (n choose k) = n! / (k! * (n - k)!) However, for the powers of 8, this would involve a lot of terms and computation, namely: (1 + 3i)^8 = 1^8 + (8 choose 1)*1^7*(3i) + (8 choose 2)*1^6*(3i)^2 + ... + (3i)^8 (1 - 3i)^8 = 1^8 + (8 choose 1)*1^7*(-3i) + (8 choose 2)*1^6*(-3i)^2 + ... + (-3i)^8 You will notice that terms with odd powers of (3i) in (1 + 3i)^8 will cancel out with the corresponding terms in (1 - 3i)^8 due to the opposite signs. Only the even powers will remain, which are identical in both expressions because (-3i)^2n = (3i)^2n for all integers n, as both will be a real number. To simplify calculation, you can just compute the terms involving even powers of i from either (1 + 3i)^8 or (1 - 3i)^8 and then double the real part to get the final answer. However, for powers of 8 and without further tricks, this could be a tedious process and is typically done using a computer algebra system. For simplicity and practicality, let me assist you in solving this with such computational help: (1 + 3i)^8 + (1 - 3i)^8 simplifies to 2 * (1^8 + (8 choose 2)*1^6*(3i)^2 + (8 choose 4)*1^4*(3i)^4 + (8 choose 6)*1^2*(3i)^6 + (3i)^8). Each individual term can be computed, keeping in mind that i^2 = -1, i^4 = i^2 * i^2 = (-1)^2 = 1, i^6 = i^4 * i^2 = 1 * (-1) = -1, and i^8 = i^6 * i^2 = (-1) * (-1) = 1. Here's the simplification of the even-i-powered terms: (8 choose 2)(3i)^2 = 28 * 9 * (-1) = -252 (8 choose 4)(3i)^4 = 70 * 81 * 1 = 5670 (8 choose 6)(3i)^6 = 28 * 729 * (-1) = -20412 (3i)^8 = 6561 * 1 = 6561 Summing these and multiplying by 2 gives the real part: 2 * (1 - 252 + 5670 - 20412 + 6561) = 2 * (1 - 252 + 5670 - 20412 + 6561) = 2 * 51568 = 103136 So the final answer is: 103136
Simplified Cube Root of a Fraction
The expression in the image is: \[ \left(\frac{2 + 2\sqrt{13}}{6}\right)^3 \] To solve this expression, you need to raise the fraction within the parentheses to the power of 3. First, simplify the fraction by dividing both the numerator and the denominator by 2, then raise the result to the third power: Simplify the fraction: \[ \left(\frac{2}{6} + \frac{2\sqrt{13}}{6}\right)^3 = \left(\frac{1}{3} + \frac{\sqrt{13}}{3}\right)^3 \] Now compute the cube of the simplified expression. Let's denote \(a = \frac{1}{3}\) and \(b = \frac{\sqrt{13}}{3}\), then the expression becomes \((a + b)^3\). Using the binomial expansion: \[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \] Apply this formula: \[ \left(\frac{1}{3} + \frac{\sqrt{13}}{3}\right)^3 = \left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right)^2\left(\frac{\sqrt{13}}{3}\right) + 3\left(\frac{1}{3}\right)\left(\frac{\sqrt{13}}{3}\right)^2 + \left(\frac{\sqrt{13}}{3}\right)^3 \] Now carry out the exponentiation and multiplication: \[ = \frac{1}{27} + 3 \cdot \frac{1}{9} \cdot \frac{\sqrt{13}}{3} + 3 \cdot \frac{1}{3} \cdot \frac{13}{9} + \frac{13\sqrt{13}}{27} \] Simplify the terms: \[ = \frac{1}{27} + \frac{\sqrt{13}}{9} + \frac{13}{9} + \frac{13\sqrt{13}}{27} \] Combine the terms: \[ = \frac{1 + 3\sqrt{13} + 39 + 13\sqrt{13}}{27} \] Combine like terms: \[ = \frac{40 + 16\sqrt{13}}{27} \] That's the simplified form of the cube of the original expression: \[ \left(\frac{2 + 2\sqrt{13}}{6}\right)^3 = \frac{40 + 16\sqrt{13}}{27} \]
Expanding and Analyzing (1 - (2/a)x)(a+x)^5
The question is asking us to show that in the expansion of \((1 - \frac{2}{a}x)(a+x)^5\), where \(a\) is a non-zero constant, the coefficient of \(x^2\) is zero. To find the coefficient of \(x^2\), we need to consider how terms which multiply together to give an \(x^2\) term could arise in the expansion. We can expand \((a+x)^5\) using the binomial theorem and then multiply the result by \((1 - \frac{2}{a}x)\). We're only interested in terms that give \(x^2\) after this multiplication, so we can ignore higher powers of \(x\). The binomial expansion of \((a + x)^5\) is given by: \((a + x)^5 = a^5 + 5a^4x + 10a^3x^2 + 10a^2x^3 + 5ax^4 + x^5\) We are interested in the terms up to \(x^2\) since multiplying by \(x^3\) or higher from the linear term outside will exceed \(x^2\). When we multiply \((1 - \frac{2}{a}x)\) by the \(a^5\), \(5a^4x\), and \(10a^3x^2\) terms, we'll get coefficients for \(x^0\), \(x^1\), and \(x^2\), respectively. Terms resulting from \(10a^2x^3\) and beyond multiplied by \(1\) would be of a higher power than \(x^2\), and are therefore irrelevant. So we focus on the following: \((1 - \frac{2}{a}x)(a^5 + 5a^4x + 10a^3x^2)\) Expanding the relevant terms: \(a^5(1 - \frac{2}{a}x) = a^5 - 2a^4x\) \(5a^4x(1 - \frac{2}{a}x) = 5a^4x - 10a^3x^2\) \(10a^3x^2(1 - \frac{2}{a}x) = 10a^3x^2 - 20a^2x^3\) Now we can add up the coefficients of \(x^2\) from these products: From \(a^5(1 - \frac{2}{a}x)\), we do not have an \(x^2\) term. From \(5a^4x(1 - \frac{2}{a}x)\), the coefficient of \(x^2\) is \(-10a^3\). From \(10a^3x^2(1 - \frac{2}{a}x)\), the coefficient of \(x^2\) is \(10a^3\). If we sum the coefficients for \(x^2\) from these two terms, we get: \(-10a^3 + 10a^3 = 0\) Hence, the coefficient of \(x^2\) in the expansion of \((1 - \frac{2}{a}x)(a+x)^5\) is zero, as required to be shown.
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