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Damm [24]
2 years ago
9

A geometric sequence is shown below.

Mathematics
1 answer:
S_A_V [24]2 years ago
8 0

a_1=2;\ a_2=-6;\ a_3=18;\ a_4=-54;\ a_5=162;\ ...

-----------------------------------------------------

A recursive rule for a geometric sequence:

a_1\\\\a_n=r\cdot a_{n-1}

r=\dfrac{a_{n+1}}{a_n}\to r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...\\\\r=\dfrac{-6}{2}=-3

Therefore \boxed{a_1=2;\qquad a_n=-3a_{n-1}}

-----------------------------------------------------

The exciplit rule:

a_n=a_1r^{n-1}

Substitute:

a_n=2(-3)^{n-1}=2(3)^n(3)^{-1}=2(3)^n\left(\dfrac{1}{3}\right)\\\\\boxed{a_n=\dfrac{2}{3}\left(-3)^n}

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The probability is \frac{56!}{64!}

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 64 \choose 8 .

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We can conclude that the probability for 8 rooks not being able to capture themselves is

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