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3 years ago

A sequence is defined by the recursive function f(n + 1) = 1/3f(n). If f(3) = 9 , what is f(1)

Mathematics

1 answer:

lord [1]3 years ago

7 0

Answer:

f(1) = 81

Step-by-step explanation:

given is f(n + 1) = 1/3f(n)

f(3) = 9

ayou need to find f(0) before you can answer what is f(1).

Just start by applying for...

n=0

f(0+1) = 1/3 * f(0)

f(1) = 1/3 * f(0)

n=1

f(1+1) = 1/3 * 1/3*f(0)

f(2) = 1/9 * f(0)

n=2

f(2+1) = 1/3 * 1/9* f(0)

f(3) = 1/27 * f(0)

But it is given that f(3) = 9 so substitute this in the line above to calculate the value of f(0)....

9 = 1/27 * f(0)

Multiply left and right of the = sing by 27 gives

f(0) = 27 * 9

f(0) = 243

We already know this next line because we started with n=0

f(1) = 1/3 * f(0)

Substitute f(0) = 243 in the line above to finally calculate the value of f(1).

f(1) = 1/3 * 243

f(1) = 81

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Step-by-step explanation:

Part 1)

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Let

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Part 2) we know that

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we have

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substitute in the formula

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The diagonal DB is equal to

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Let

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substitute in the formula

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step 5

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$324\pi\ units^{2}-648\ units^{2}+162\pi\ units^{2}=(486\pi-648)\ units^{2}$

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