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den301095 [7]
3 years ago
15

If f(1)=3 and f(n)=-2f(n-1)+1,then f(5)=

Mathematics
1 answer:
Drupady [299]3 years ago
5 0
F(5) = - 2f(4) + 1
f(4) = -2f(3) + 1
f(3) = -2f(2) + 1
f(2) = -2f(1) + 1
Therefore:
 f(2) = -2(3) + 1 = -5
f(3) = -2(-5) + 1 = 11
f(4) = -2(11) + 1 = -21
Therefore f(5) = -2(-21) + 1 = 43
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Also, let \%R_{A} be the quantity of nonzero elements in R_{A}.

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Case 2: \%R_{A}=2 and \%R_B=4. There are 3 ways to position the -1 in R_A, 2 ways to put the remaining -1 in R_B (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of \%R_{A}=4,\%R_{B}=2 for a complete of 24 ways.

Case 3: \%R_A=\%R_B=2. There are 3 ways to put the -1 in R_{A}. Now, there are two cases on what happens next.

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  • The 1 in R_B doesn't go directly under the -1 in R_A. There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

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