All categories

4 years ago

30 POINTS!!!!!!

Mathematics

2 answers:

sergeinik [125]4 years ago

8 0

Answer with explanation:

→Infinite Geometric Series

$a_{1}=960\\\\r=\frac{1}{4}$

→The geometric series having common ratio r, and first term a,can be written as: $a, ar,ar^2,ar^3,ar^4,.......$

→So, the geometric Series can be Written as:

$960, 960 *\frac{1}{4},960*[\frac{1}{4}]^2,960*[\frac{1}{4}]^3,......\\\\ 960,240,60,15,.....$

→Sum of Infinite geometric Series

=960+240+60+15+.......

$={S_{\text{Infinity}}=\sum_{n=1}^{\infty }960*r^{n-1}=\frac{a}{1-r}$

$=\frac{960}{1-\frac{1}{4}}\\\\=\frac{960*4}{3}\\\\=320*4=1280\\\\=\frac{a}{1-r}$

Sum ,to infinity, Which is upper limit of this population=1280

sergey [27]4 years ago

4 0

Answer: The sum that will be the upper limit of this population is 1280.

Step-by-step explanation:

Since we have given that

Initial population a₁ = 960

Common ratio = $\frac{1}{4}$

So, We have to write the sum in sigma notation:

$\sum (ar^{n-1})\\\\=\sum 960(\frac{1}{4})^{n-1}\\\\$

Since $r=\frac{1}{4}$

so, the sum is convergent, then,

$\sum 960(\frac{1}{4})^{n-1}=\frac{a}{1-r}=\frac{960}{1-\frac{1}{4}}=\frac{960}{\frac{3}{4}}=\frac{960\times 4}{3}=320\times 4=1280$

Hence, the sum that will be the upper limit of this population is 1280.

You might be interested in

Are 15:35 and 25:45 the same thing]

marishachu [46]

They are not the same thing.

3 0

3 years ago

Help plz nd thank you<br> (true or false questions)<br> 1.) -19-2+54<12x97+-40<br> 2.) x+1=1+x

Anastaziya [24]

Number one is 33 < 1118 so true and number two is infinite solutions so it would be true as well

6 0

3 years ago

Note: We can only use the expanded power rule when there isn't any addition or subtraction. In the example below you need to mul

Simora [160]

21 hope this helps and your welcome

4 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago

I need help with this yes or no picking?

Ivenika [448]

Use photomath. Do not click on the link the other person which is actually a bot put.

6 0

3 years ago