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BaLLatris [955]
3 years ago
7

Find the solution of the recurrence relation an = 3an−1 −3an−2 +an−3 if a0 = 2, a1 = 2, and a2 = 4.

Mathematics
1 answer:
Shkiper50 [21]3 years ago
5 0

Using the recurrence relation, we can find a couple more values in the sequence:

  • a3 = 3a2 -3a1 +a0 = 3(4) -3(2) +2 = 8
  • a4 = 3a3 -3a2 +a1 = 3(8) -3(4) +2 = 14

First differences are 0, 2, 4, 6, ...

Second differences are constant at 2, so the function is quadratic.

The sequence can be described by the quadratic ...

... an = n² -n +2

_____

We know the value for n=0 is 2, so we can find <em>a</em> and <em>b</em> using the given values for a1 and a2.

... an = an² +bn +2

... a1 = 2 = a·1² +b·1 +2 . . . . for n=1

... a + b = 0

... a2 = 4 = a·2² -a·2 +2 . . . . for n = 2; using b=-a from the previous equation

... 2 = 2a

... a = 1 . . . . so b = -1

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What is the remainder when f(x) = x^2 + 14x − 8 is divided by (x − 5)? 103 88 87 72
il63 [147K]

Answer:

<h2>The answer is 87</h2>

Step-by-step explanation:

f(x) = x² + 14x - 5

To find the reminder when f(x) is divided by x - 5 , substitute the value of x into the above formula

That's

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x = 5

So we have

f(5) = 5² + 14(5) - 8

f(5) = 25 + 70 - 8

f(5) = 95 - 8

We have the final answer as

<h3>87</h3>

Hope this helps you

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