If f(1)=1,f(2)=0 and f(n)=f(n-1)+2f(n-2) then find the value of f(4). Answer:

Calculate f(3): To find f(4), we need to first find f(3) using the given recursive formula f(n)=f(n-1)+2f(n-2). We know f(2)=0 and f(1)=1, so we can calculate f(3) as follows: f(3) = f(3-1) + 2f(3-2) = f(2) + 2f(1) = 0 + 2(1) = 2Find f(4): Now that we have f(3), we can use it to find f(4) using the same recursive formula: f(4) = f(4-1) + 2f(4-2) = f(3) + 2f(2) = 2 + 2(0) = 2

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If
f(1)=1,f(2)=0 and
f(n)=f(n-1)+2f(n-2) then find the value of
f(4).
Answer:

If $f(1)=1, f(2)=0$ and $f(n)=f(n-1)+2 f(n-2)$ then find the value of $f(4)$ . $\newline$ Answer:

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Precalculus

Find the roots of factored polynomials

Full solution

Q. If

f(1)=1, f(2)=0

and

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

then find the value of

f(4)

\newline

Answer:

Calculate $f(3)$ : To find $f(4)$ , we need to first find $f(3)$ using the given recursive formula $f(n)=f(n-1)+2f(n-2)$ . We know $f(2)=0$ and $f(1)=1$ , so we can calculate $f(3)$ as follows: $f(3) = f(3-1) + 2f(3-2)$ $= f(2) + 2f(1)$ $= 0 + 2(1)$ $= 2$
Find $f(4)$ : Now that we have $f(3)$ , we can use it to find $f(4)$ using the same recursive formula: $\newline$ $f(4) = f(4-1) + 2f(4-2)$ $\newline$ $= f(3) + 2f(2)$ $\newline$ $= 2 + 2(0)$ $\newline$ $= 2$