Find the sum of each of the following series:, №18010983, 15.11.2022 04:48

Mathematics

Find the sum of each of the following series: (a) 2 + 7 + 12 + +92

Step-by-step answer

P Answered by Master

Arithmetic SequenceWe can observe that we are adding 5 each time to the numbers, meaning this is an arithmetic sequence with a common difference d=5Explicit RelationWe can establish the relation,u(n)=u⁰+nd

where u(n) is the nth term (any term).

u⁰ is out first term (2)

d is the common difference (5)

u(n)=u⁰+nd92 =2 +5n5n =90n =18This sequence admits 19 terms since we start counting from 0 till 18.SummationS= number of terms/2 ×(u¹⁸+u⁰) $s = \frac{19}{2} \times (92 + 2)$

s = 9.5(94)

s = 893

That's ur answer :)

Mathematics

2arithmetic sequences and series find the indicated term of each arithmetic sequence. 1. find the sixtieth term of the arithmetic sequence if a1 = 418 and d = 12. 2. find a23 in the sequence, –18, –34, –50, –66, write an equation for the nth term of each arithmetic sequence. 3. 45, 30, 15, 0, 4. –87, –73, –59, –45, find the sum of each arithmetic series. 5. 5 + 7 + 9 + 11 + + 27 6. –4 + 1 + 6 + 11 + + 91 7. 13 + 20 + 27 + + 272 8. 89 + 86 + 83 + 80 + + 20 9. ∑ (1 − 2n) 4 n=1 10. ∑ (5 + 3j) 6 j=1 11. ∑ (9 − 4n) 5 n=1 12. ∑ (2k + 1) 10 k=4 13. ∑

Step-by-step answer

P Answered by Specialist

8

I have the first 8 answers for you:
1: 1126
2: -370
3: a_n=45-15(n-1)

4:

5: 192
6: 870
7: 5414
8: 1308

The formula for #1 would be a_n=418+12(n-1)

. Using 60 for n, we have
418+12(60-1) = 1126

The formula for #2 would be a_n=-18-16(n-1)

. Using 23 for n, we have
-18-16(23-1) = -370

The general form for this sequence is a_n=a_1+d(n-1)

, where a₁ is the first term and d is the common difference. For #3, the first term is 45 and the common difference is -15. For #4, the first term is -87 and the common difference is 14.

For #5-8, add together the terms.

Mathematics

PLZ HELP! 20 POINTS! A student writes the arithmetic series 8 + 13 + ... + 43 in summation notation (pictured below) A) Describe the error. B) Find the sum of the arithmetic series C) Write the explicit and recursive formulas for each term in the series. Be sure to label them. I also attached the entire question below.

Step-by-step answer

P Answered by PhD

7

a) The error is that, the initial value is n=1 NOT n=3

b) The sum is $a_n=a_{n+1}+5=192$

c)The explicit formula is a_n=5n+3

The recursive formula is $a_n=a_{n+1}+5$ ,

Step-by-step explanation:

The given arithmetic series is 8 + 13 + ... + 43.

The first term is a_1=8 , the common difference is d=13-8=5

The nth term is given by:

a_n=a_1+d(n-1)

We substitute the values to get:

$a_n=8+5(n-1)\\a_n=8+5n-5\\\\a_n=3+5n$

To find how many terms are in the sequence we solve the equation:

$3+5n=43\\\implies 5n=43-3\\5n=40\\n=8$

The summation notation is $\sum_{n=1}^8(3+5n)$

The error the student made is in the initial value.

It should be n=1 NOT n=3

b) The sum of the arithmetic series is calculated using:

$S_n=\frac{n}{2}(a+l)$

We substitute o get:

$S_8=\frac{8}{2}(5+43)$

S_8=4(48)

S_8=192

c) The explicit formula we already calculated in a), which is a_n=3+5n

The recursive formula is given as:

$a_n=a_{n+1}+d$

We substitute d=5 to get:

$a_n=a_{n+1}+5$

Mathematics

40 points! A Student write the arithmetic series 8 + 13 + … + 43 in summation notation as (the attachment shown). a): Describe the error. b): Find the sum of the arithmetic series. c): Write the explicit and recursive formulas for each term in the series. Be sure to label them.

Step-by-step answer

P Answered by PhD

1

I'm terribly sorry but this i cannot answer correctly for you

Step-by-step explanation:

Mathematics

find a formula for the nth partial sum of each series 4+(4/3)+(4/9) and use it to find the series’ sum of the series converges.

Step-by-step answer

P Answered by PhD

We can see that:

a₀ = 4

a₁ = 4/3

a₂ = 4/9.

We easily can see that the n-th therm of this will be:

aₙ = 4/3^n.

Now, the denominator increases as n increases, then for a really large n, aₙ will tend asymptotically to zero. This means that this summation converges.

We can write this summation as:

4*∑(1/3)^n.

The sum for the first N-th therms is:

We know that for a summation:

1 + r + r^2 + r^3 + ... + r^N = (1 - r^(N + 1))/(1 - r)

4 + 4*(1/3) + 4*(1/3)^2 + ... + 4*(1/3)^N = 4*( 1 - (1/3)^(N + 1))/(1 - (1/3))

Now, for the complete sum we have that:

n = {0, 1, 2, ...}

We know that for a summation:

∑a*r^n

with n = {0, 1, ...}

the sum is = a/(1 + r).

in this case we have:

a = 4, r = 1/3.

Then the sum is:

S = 4/(1 - 1/3) = 4/(2/3) = 3*2 = 6

Mathematics

PLZ HELP! 20 POINTS! A student writes the arithmetic series 8 + 13 + ... + 43 in summation notation (pictured below) A) Describe the error. B) Find the sum of the arithmetic series C) Write the explicit and recursive formulas for each term in the series. Be sure to label them. I also attached the entire question below.

Step-by-step answer

P Answered by PhD

7

a) The error is that, the initial value is n=1 NOT n=3

b) The sum is $a_n=a_{n+1}+5=192$

c)The explicit formula is a_n=5n+3

The recursive formula is $a_n=a_{n+1}+5$ ,

Step-by-step explanation:

The given arithmetic series is 8 + 13 + ... + 43.

The first term is a_1=8 , the common difference is d=13-8=5

The nth term is given by:

a_n=a_1+d(n-1)

We substitute the values to get:

$a_n=8+5(n-1)\\a_n=8+5n-5\\\\a_n=3+5n$

To find how many terms are in the sequence we solve the equation:

$3+5n=43\\\implies 5n=43-3\\5n=40\\n=8$

The summation notation is $\sum_{n=1}^8(3+5n)$

The error the student made is in the initial value.

It should be n=1 NOT n=3

b) The sum of the arithmetic series is calculated using:

$S_n=\frac{n}{2}(a+l)$

We substitute o get:

$S_8=\frac{8}{2}(5+43)$

S_8=4(48)

S_8=192

c) The explicit formula we already calculated in a), which is a_n=3+5n

The recursive formula is given as:

$a_n=a_{n+1}+d$

We substitute d=5 to get:

$a_n=a_{n+1}+5$

Mathematics

40 points! A Student write the arithmetic series 8 + 13 + … + 43 in summation notation as (the attachment shown). a): Describe the error. b): Find the sum of the arithmetic series. c): Write the explicit and recursive formulas for each term in the series. Be sure to label them.

Step-by-step answer

P Answered by PhD

1

I'm terribly sorry but this i cannot answer correctly for you

Step-by-step explanation:

Mathematics

2arithmetic sequences and series find the indicated term of each arithmetic sequence. 1. find the sixtieth term of the arithmetic sequence if a1 = 418 and d = 12. 2. find a23 in the sequence, –18, –34, –50, –66, write an equation for the nth term of each arithmetic sequence. 3. 45, 30, 15, 0, 4. –87, –73, –59, –45, find the sum of each arithmetic series. 5. 5 + 7 + 9 + 11 + + 27 6. –4 + 1 + 6 + 11 + + 91 7. 13 + 20 + 27 + + 272 8. 89 + 86 + 83 + 80 + + 20 9. ∑ (1 − 2n) 4 n=1 10. ∑ (5 + 3j) 6 j=1 11. ∑ (9 − 4n) 5 n=1 12. ∑ (2k + 1) 10 k=4 13. ∑

Step-by-step answer

P Answered by Master

8

I have the first 8 answers for you:
1: 1126
2: -370
3: a_n=45-15(n-1)

4:

5: 192
6: 870
7: 5414
8: 1308

The formula for #1 would be a_n=418+12(n-1)

. Using 60 for n, we have
418+12(60-1) = 1126

The formula for #2 would be a_n=-18-16(n-1)

. Using 23 for n, we have
-18-16(23-1) = -370

The general form for this sequence is a_n=a_1+d(n-1)

, where a₁ is the first term and d is the common difference. For #3, the first term is 45 and the common difference is -15. For #4, the first term is -87 and the common difference is 14.

For #5-8, add together the terms.

Mathematics

find a formula for the nth partial sum of each series 4+(4/3)+(4/9) and use it to find the series’ sum of the series converges.

Step-by-step answer

P Answered by PhD

We can see that:

a₀ = 4

a₁ = 4/3

a₂ = 4/9.

We easily can see that the n-th therm of this will be:

aₙ = 4/3^n.

Now, the denominator increases as n increases, then for a really large n, aₙ will tend asymptotically to zero. This means that this summation converges.

We can write this summation as:

4*∑(1/3)^n.

The sum for the first N-th therms is:

We know that for a summation:

1 + r + r^2 + r^3 + ... + r^N = (1 - r^(N + 1))/(1 - r)

4 + 4*(1/3) + 4*(1/3)^2 + ... + 4*(1/3)^N = 4*( 1 - (1/3)^(N + 1))/(1 - (1/3))

Now, for the complete sum we have that:

n = {0, 1, 2, ...}

We know that for a summation:

∑a*r^n

with n = {0, 1, ...}

the sum is = a/(1 + r).

in this case we have:

a = 4, r = 1/3.

Then the sum is:

S = 4/(1 - 1/3) = 4/(2/3) = 3*2 = 6

Mathematics

As of 2010, the new york yankees had one more world series championships than any other team. in fact, the yankees had won 3 fewer than 3 times the number of world series championships won by the second most winning team, the st. louis cardinals. the sum of the two teams world series championships is 37. how many times has each team won the world series? solve using a system of equations. please show work.

Step-by-step answer

P Answered by Master

3

Step-by-step explanation:

first you get your equations, because the game sums are based upon the st louis cardinals we will use them as x.

According to the problem the yankees won 3 fewer (-3) that 3 times (3x) the st louis cardinals which brings their equation to 3x-3

The sum of these to equations equals 37

37=x+3x-3

when you solve, you get x=10

then you plug it into the DIFERENT equations

3(10)-3=27 (yankees)

x=10

then check using your preffered method

Find the sum of each of the following series:
(a) 2 + 7 + 12 + +92

Step-by-step answer

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