All categories

3 years ago

What's the explicit formula for the sequence :")

Mathematics

1 answer:

zzz [600]3 years ago

7 0

Hello from MrBillDoesMath!

Answer:

a(n) = 12 - 3(n-1) , n = 1,2,3.,,

Discussion:

The terms are 12, 9, 6, 3 so we are apparently subtracting 3 from each succeeding term. . Hence

a(n) = 12 - 3(n-1) , n = 1,2,3.,,

Thank you,

MrB

You might be interested in

Describe the pattern below<br><br> 3 6.5 11 16.5 23

ipn [44]

6.5 - 3 = 3.5
11 - 6.5 = 4.5
16.5 - 11 = 5.5
23 - 16.5 = 6.5

From this, you can see that the number being added is increasing by 1 each time.

Therefore, the pattern is adding 3.5, then 4.5, then 5.5, then 6.5, and so on.

5 0

3 years ago

A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e

My name is Ann [436]

Answer:

$\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)} \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

Step-by-step explanation:

If the system contains $n$ atoms, we can arrange $r$ quanta of energy in

$\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$

ways.

$\mathbf{a)}$

In this case,

$n = 2, r=1$ .

Therefore,

$\binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2$

which means that we can arrange 1 quanta of energy in 2 ways.

$\mathbf{b)}$

In this case,

$n = 20, r=10$ .

Therefore,

$\binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756$

which means that we can arrange 10 quanta of energy in 184 756 ways.

$\mathbf{c)}$

In this case,

$n = 2 \times 10^{23}, r = 10^{23}$ .

Therefore, we obtain that the number of ways is

$\binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$