If f(1)=3, and f(n+1)-f(n)=3(4^n-1), then for | vedclass

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(C) Given $f(1)=3$ and $f(n+1)-f(n)=3(4^n-1)$.We can write the recurrence as a telescoping sum:$f(n) = f(1) + \sum_{k=1}^{n-1} (f(k+1)-f(k))$$f(n) = 3

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$4^n-3n+2$

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(C) Given $f(1)=3$ and $f(n+1)-f(n)=3(4^n-1)$.We can write the recurrence as a telescoping sum:$f(n) = f(1) + \sum_{k=1}^{n-1} (f(k+1)-f(k))$$f(n) = 3 + \sum_{k=1}^{n-1} 3(4^k-1)$$f(n) = 3 + 3 \left( \sum_{k=1}^{n-1} 4^k - \sum_{k=1}^{n-1} 1 \right)$Using the geometric series sum formula $\sum_{k=1}^{n-1} 4^k = \frac{4(4^{n-1}-1)}{4-1} = \frac{4^n-4}{3}$:$f(n) = 3 + 3 \left( \frac{4^n-4}{3} - (n-1) \right)$$f(n) = 3 + (4^n-4) - 3(n-1)$$f(n) = 3 + 4^n - 4 - 3n + 3$$f(n) = 4^n - 3n + 2$

If $f(1)=3$, and $f(n+1)-f(n)=3(4^n-1)$, then for all $n \in N$, $f(n)=$

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