What are the last three terms of the binomial expansion 32x³ + 80x'y + 80x³y² +
1 answer:
The last three terms of the given binomial expansion are 40x²y³ + 10xy⁴ + y⁵
<h3>Binomial expansion</h3>From the question, we are to determine the last three terms of the binomial expansion 32x⁵ + 80x⁴y + 80x³y² + __ + __ + __
The highest power of the expansion is 5
Using the binomial expansion theorem
(m +n)⁵ = m⁵ + 5m⁴n + 10m³n² + 10m²n³ + 5mn⁴ + n⁵
Now, we will determine the values of m and n by comparison
By comparison,
m⁵ = 32x⁵
m⁵ = (2x)⁵
∴ m = 2x
Also,
5m⁴n = 80x⁴y
Divide both sides by 5, to get
m⁴n = 16x⁴y
Put m = 2x into the equation
(2x)⁴n = 16x⁴y
16x⁴n = 16x⁴y
∴ n = y
Thus, the expression being expanded is (2x + y)⁵
By binomial expansion, the expansion of (2x + y)⁵ is
(2x + y)⁵ = (2x)⁵ + 5(2x)⁴y + 10(2x)³(y)² + 10(2x)²(y)³ + 5(2x)(y)⁴ + y⁵
(2x + y)⁵ = 32x⁵ + (5×16)x⁴y + (10×8)x³(y)² + (10×4)x²(y)³ + (5×2)x(y)⁴ + y⁵
(2x + y)⁵ = 32x⁵ + 80x⁴y + 80x³y² + 40x²y³ + 10xy⁴ + y⁵
Hence, the last three terms of the given binomial expansion are 40x²y³ + 10xy⁴ + y⁵
Learn more on Binomial expansion here: brainly.com/question/13602562
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the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
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Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.
On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057