Example Question - coefficient of x^2
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Expanding Binomial with Coefficients of x² and x³
The image shows several mathematical problems, but I will focus on question number 2, as requested: "2. The coefficient of x^2 and x^3 in the expansion of (3-2x)^5 are a and b respectively. Find the value of p = a-b." To find the coefficients of x² and x³ in the expansion of (3-2x)⁵, we can use the binomial theorem. The binomial theorem states that (a + b)ⁿ = Σ from k = 0 to n (n choose k) * a^(n-k) * b^k. For (3 - 2x)⁵, we have a = 3 and b = -2x. Now we need to find the terms where the exponent of x is 2 and 3. The general term in binomial expansion is T(k+1) = (n choose k) * a^(n-k) * b^k. For the x² term (k = 2): T(3) = (5 choose 2) * 3^(5-2) * (-2x)² T(3) = 10 * 3³ * 4x² T(3) = 10 * 27 * 4x² T(3) = 1080 x² So, a = 1080. For the x³ term (k = 3): T(4) = (5 choose 3) * 3^(5-3) * (-2x)³ T(4) = 10 * 3² * -8x³ T(4) = 10 * 9 * -8x³ T(4) = -720 x³ So, b = -720. Finally, we have to find p = a - b: p = a - b p = 1080 - (-720) p = 1080 + 720 p = 1800. Therefore, the value of p is 1800.
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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