Question

Enter the enter the first four terms of the sequence defined by the given rule assume that the domain of each function is the set of whole numbers greater than f(1) =4, f(n)= (-5) •F(n-1)+11 The first 4 terms of the sequences are?

Answer 1

Answer:

  {1, -9, 56, -269}

Step-by-step explanation:

Evaluate the rule for n=2, 3, and 4 in sequence.

For n=2

  f(2) = (-5)f(1) +11 = (-5)(4) +11 = -9

  f(3) = (-5)f(2) +11 = (-5)(-9) +11 = 56

  f(4) = (-5)f(3) +11 = (-5)(56) +11 = -269

The first four terms of the sequence are {1, -9, 56, -269}.

Answer 2

The first four terms of the sequence are 4, -9. 56 and -269

Given the sequence defined by the given rule $f(n)= (-5) \cdot f(n-1)+11$

If the first term f(1) = 4.

Get the second term;

f(2) = -5f(2-1) + 11

f(2) = -5f(1) + 11

f(2) = -5(4) + 11

f(2) = -20 + 11

f(2) = -9

Get the third term;

f(3) = -5f(3-1) + 11

f(3) = -5f(2) + 11

f(3) = -5(-9) + 11

f(3) = 45 + 11

f(3) = 56

Get the fourth term;

f(4) = -5f(4-1) + 11

f(4) = -5f(3) + 11

f(4) = -5(56) + 11

f(4) = -280 + 11

f(4) = -269

Hence the first four terms of the sequence are 4, -9. 56 and -269

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