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vlabodo [156]
2 years ago
13

1. Which formula defines the sequence f(1)=2, f(2)= 6, f(3)= 10, f(4)= 14, f(5)= 18?

Mathematics
1 answer:
Natasha_Volkova [10]2 years ago
7 0

Answer:

f(n) =4 + f(n-1) \\for\ n>2

Step-by-step explanation:

Given

f(1)=2\\ f(2)= 6\\ f(3)= 10\\ f(4)= 14\\ f(5)= 18

Required

Determine the formula

First, we need to solve common difference (d)

d = f(n) - f(n-1)

Take n as 2

d = f(2) - f(2-1)

d = 6 - 2

d = 4

Represent each function as a sum of the previous

f(1) = 2

f(2) = 2 + 4 = f(1) + 4

f(3) = 6 + 4 = f(2) + 4

f(4) = 10 + 4 = f(3) + 4

f(5) = 14 + 4 = f(4) + 4

Represent the function as f(n)

f(n) =f(n-1) + 4

Reorder

f(n) =4 + f(n-1) \\for\ n>2

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Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫
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\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1

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###

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which gives

x(t)^{2/3}+y(t)^{2/3}=(4\cos^3t)^{2/3}+(4\sin^3t)^{2/3}=4^{2/3}(\cos^2t+\sin^2t)=4^{2/3}

as needed. Then with 0\le t\le2\pi, we compute the area via Green's theorem using the same setup as before:

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