regents question #24
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pka
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\(\displaystyle f(2)=-2 * f(2-1)+1=~?\).
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It was worked out by working it!
You are given a value for f when x = 1, and you are given a formula for finding the value of f for one more than a given value of x (that is, the value of f for x + 1). Use this to find the value of f(2).
Then use this result to find the value of f(3). Then find the value of f(4). And then f(5).
Different way, same result:
f(n) = -2 f(n-1) + 1
= -2 (- 2 f(n-2) + 1) + 1 = (-2)2 f(n-2) + (-2)1 + 1
= -2 ( (-2)2 f(n-2) + (-2) + 1 ) + 1 = (-2)3 f(n-3) + (-2)2 + (-2)1 + 1
= ...
= -2 ( (-2)j-1 f(n-j) + (-2)j-2 + ... + (-2)1 + 1 ) + 1 = (-2)j f(n-j) + (-2)j-1 + ... + (-2)2 + (-2)1 + 1
= ...
= (-2)n-1 f(1) + (-2)n-2 + ... + (-2)2 + (-2)1 + 1
= (-2)n-1 3 + \(\displaystyle \Sigma_{j=0}^{j=n-2} (-2)^j\)
Now you have the sum of a geometric series which, hopefully, you know how to compute so do so and compute f(5).
Or if you don't know the sum of a geometric series highlight the following between the >>>>> and <<<<<
>>>>>
= 3 * (-2)n-1 + \(\displaystyle \frac{(-2)^{n-1} - 1}{-2 - 1}\)
= 3 * (-2)n-1 - [ (-2)n-1 - 1 ] / 3
Now compute f(5).
<<<<<
Hello, Roepap!
Do you understand: \(\displaystyle f(n) \:=\:-2f(n-1) + 1\) ?
It says: The \(\displaystyle n^{th}\) term is -2 times the preceding term, plus 1.
We are given: \(\displaystyle f(1) = 3\)
Then: .\(\displaystyle f(2)\:=\:\text{-}2\cdot f(1) + 1 \:=\:\text{-}2(3) + 1 \quad\Rightarrow\quad f(2)\:=\:\text{-}5\)
Then: .\(\displaystyle f(3) \:=\:\text{-}2\cdot f(2) + 1 \:=\:\text{-}2(\text{-}5)+1 \quad\Rightarrow \quad f(3) \:=\:11\)
Then: .\(\displaystyle f(4) \:=\:\text{-}2\cdot f(3) + 1 \:=\:\text{-}2(11)+1 \quad\Rightarrow \quad f(4) \:=\:\text{-}21\)
Then: .\(\displaystyle f(5)\:=\:\text{-}2\cdot f(4)+1 \:=\:\text{-}2(\text{-}21) + 1 \quad\Rightarrow\quad f(5) \:=\:43\)
Do you understand: \(\displaystyle f(n) \:=\:-2f(n-1) + 1\) ?
It says: The \(\displaystyle n^{th}\) term is -2 times the preceding term, plus 1.
We are given: \(\displaystyle f(1) = 3\)
Then: .\(\displaystyle f(2)\:=\:\text{-}2\cdot f(1) + 1 \:=\:\text{-}2(3) + 1 \quad\Rightarrow\quad f(2)\:=\:\text{-}5\)
Then: .\(\displaystyle f(3) \:=\:\text{-}2\cdot f(2) + 1 \:=\:\text{-}2(\text{-}5)+1 \quad\Rightarrow \quad f(3) \:=\:11\)
Then: .\(\displaystyle f(4) \:=\:\text{-}2\cdot f(3) + 1 \:=\:\text{-}2(11)+1 \quad\Rightarrow \quad f(4) \:=\:\text{-}21\)
Then: .\(\displaystyle f(5)\:=\:\text{-}2\cdot f(4)+1 \:=\:\text{-}2(\text{-}21) + 1 \quad\Rightarrow\quad f(5) \:=\:43\)