Find the first five terms of the sequence in which a1 = –10 and an = 4an – 1 + 7, if n ≥ 2.

The new sequence

-40, -160, -640, -2560, -10240, -40960

Step-by-step explanation:

It is given that,

aₙ = 4a₍ₙ₋₁₎ and a₁ = -10

To find the first 7 terms of the sequence

a₁ = -10

a₂ = 4a₁ = 4 * (-10) = -40

a₃ = 4a₂ = 4 * (-40) = -160

a₄ = 4a₃ = 4 * ( -160) = -640

a₅ = 4a₄ = 4 * (-640) = -2560

a₆ = 4a₅ = 4 * (-2560) = -10240

a₇ = 4a₆ = 4 * (-10240) = -40960

Therefore the new sequence

-40, -160, -640, -2560, -10240, -40960

Given Below

Step-by-step explanation:

Given

 for n greater than equal to 2

using above equation

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

a

step-by-step explanation: