If f(1)=2,f(2)=1 and f(n)=3f(n-1)+f(n-2) then find the value of f(6). Answer:

Initialize Recursive Formula: To find the value of f(6), we need to use the recursive formula f(n) = 3f(n-1) + f(n-2) and the initial conditions f(1) = 2 and f(2) = 1. We will calculate the values of f(3), f(4), f(5), and then f(6) step by step.Calculate f(3): First, we find f(3) using the recursive formula: f(3) = 3f(2) + f(1) = 3(1) + 2 = 3 + 2 = 5. So, f(3) = 5.Calculate f(4): Next, we find f(4) using the recursive formula: f(4) = 3f(3) + f(2) = 3(5) + 1 = 15 + 1 = 16. So, f(4) = 16.Calculate f(5): Then, we find f(5) using the recursive formula: f(5) = 3f(4) + f(3) = 3(16) + 5 = 48 + 5 = 53. So, f(5) = 53.Calculate f(6): Finally, we find f(6) using the recursive formula: f(6) = 3f(5) + f(4) = 3(53) + 16 = 159 + 16 = 175. So, f(6) = 175.

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Precalculus

Find the roots of factored polynomials

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Q. If

f(1)=2, f(2)=1

and

f(n)=3 f(n-1)+f(n-2)

then find the value of

f(6)

\newline

Answer:

Initialize Recursive Formula: To find the value of $f(6)$ , we need to use the recursive formula $f(n) = 3f(n-1) + f(n-2)$ and the initial conditions $f(1) = 2$ and $f(2) = 1$ . We will calculate the values of $f(3)$ , $f(4)$ , $f(5)$ , and then $f(6)$ step by step.
Calculate $f(3)$ : First, we find $f(3)$ using the recursive formula: $\newline$ $f(3) = 3f(2) + f(1) = 3(1) + 2 = 3 + 2 = 5.$ $\newline$ So, $f(3) = 5.$
Calculate $f(4)$ : Next, we find $f(4)$ using the recursive formula: $\newline$ $f(4) = 3f(3) + f(2) = 3(5) + 1 = 15 + 1 = 16.$ $\newline$ So, $f(4) = 16.$
Calculate $f(5)$ : Then, we find $f(5)$ using the recursive formula: $\newline$ $f(5) = 3f(4) + f(3) = 3(16) + 5 = 48 + 5 = 53.$ $\newline$ So, $f(5) = 53.$
Calculate $f(6)$ : Finally, we find $f(6)$ using the recursive formula: $\newline$ $f(6) = 3f(5) + f(4) = 3(53) + 16 = 159 + 16 = 175.$ $\newline$ So, $f(6) = 175.$