Find the sum of the first 8 terms of the sequence: -5. 15, -45, 135, .......

1 answer:

7 0

Sum of geometric sequence

for the sum where the initial value is a₁ and the common ratio is r and the term is n

$S_n= \frac{a_1(1-r^n)}{1-r}$

common ratio is -3
-5 times -3 is 15, -15 times -3=-45 etc
first term is -5
and we want the 8th term

$S_8= \frac{-5(1-(-3)^8)}{1-(-3)}$
$S_8= \frac{-5(1-6561)}{1+3}$
$S_8= \frac{-5(-6560)}{4}$
$S_8= \frac{32800}{4}$
$S_8= 8200$

the sum is 8200

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Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined: