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f(n)=93+4(n-1)
Complete the recursive formula of 
f(n).

{:[f(1)=],[f(n)=f(n-1)+]:}

f(n)=93+4(n1) f(n)=93+4(n-1) \newlineComplete the recursive formula of f(n) f(n) .\newlinef(1)=f(n)=f(n1)+ \begin{array}{l} f(1)=\square \\ f(n)=f(n-1)+\square\\ \end{array}

Full solution

Q. f(n)=93+4(n1) f(n)=93+4(n-1) \newlineComplete the recursive formula of f(n) f(n) .\newlinef(1)=f(n)=f(n1)+ \begin{array}{l} f(1)=\square \\ f(n)=f(n-1)+\square\\ \end{array}
  1. Establish Base Case: To find the recursive formula for the sequence, we need to express f(n)f(n) in terms of f(n1)f(n-1). The given explicit formula is f(n)=93+4(n1)f(n)=93+4(n-1). Let's first find f(1)f(1) to establish the base case of the recursive formula.\newlineCalculation: f(1)=93+4(11)=93+4(0)=93+0=93f(1) = 93 + 4(1-1) = 93 + 4(0) = 93 + 0 = 93
  2. Find f(2)f(2): Now, let's find f(2)f(2) using the explicit formula to see the relationship between f(2)f(2) and f(1)f(1).\newlineCalculation: f(2)=93+4(21)=93+4(1)=93+4=97f(2) = 93 + 4(2-1) = 93 + 4(1) = 93 + 4 = 97
  3. Identify Recursive Relationship: We can see that f(2)f(2) is 44 more than f(1)f(1). This is because the sequence increases by 44 each time (as indicated by the +4+4 in the explicit formula). Therefore, the recursive formula will involve adding 44 to the previous term.\newlineCalculation: f(n)=f(n1)+4f(n) = f(n-1) + 4
  4. Recursive Formula: We have established the base case and the recursive step. The recursive formula for the sequence is:\newlinef(1)=93f(1) = 93\newlinef(n)=f(n1)+4f(n) = f(n-1) + 4 for n > 1

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