Expand and evaluate the series n=5 ai=2×i^2+3 (2023)

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23.05.2023the following question

Expand and evaluate the series n=5 ai=2×i^2+3 (1) 0

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23.05.2023, solved by verified expert

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Mathematics

Expand and evaluate the series n=5 ai=2×i^2+3

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P Answered by PhD

Hello,

Expand and evaluate the series n=5 ai=2×i^2+3 (7)

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Mathematics

5. evaluate the series 50 + 10 + 2 + . . 62 the series diverges; it does not have a sum. 62.5

Step-by-step answer

P Answered by PhD Expand and evaluate the series n=5 ai=2×i^2+3 (8) 7

62.5

Step-by-step explanation:

By dividing 50 by 10 and 10 by 2 we can find that r, or the common ratio is 1/5, meaning that this series converges, so it is solvable. The formula for a convergent series is Expand and evaluate the series n=5 ai=2×i^2+3 (9). By plugging in 1/5 for r and 50 for Expand and evaluate the series n=5 ai=2×i^2+3 (10) we can find that the sum of this infinite series is 62.5

Mathematics

the series has 8 terms.evaluate the series 1/2,3/2,5/2,…,15/2 you

Step-by-step answer

P Answered by PhD Expand and evaluate the series n=5 ai=2×i^2+3 (11) 1

This is an arithmetic sequence, not a series, (series are infinite, sequences are finite)because each term is a constant difference from the term preceding it, called the common difference.

Technically the sum of any arithmetic sequence is:

s(n)=(2an+dn^2-dn)/2

However the above equation is derived from the fact that the sum of an arithmetic sequence is the average of the first and last terms times the number of terms in the sequence.

All we really have to "solve" for then is the number of terms in the sequence...the rule for the sequence of an arithmetic sequence is:

a(n)=a+d(n-1), a=first term, d=common difference, n=term number. Since we know a=1/2 and d=1 we can solve for n...

a(n)=1/2+n-1

a(n)=n-1/2, now we just solve for n when a(n)=15/2

15/2=n-1/2

n=16/2

n=8 so there are seven terms, now we can say:

s(8)=8(1/2+15/2)/2

s(8)=32

Mathematics

The first five terms of a sequence are 2, 5/3, 3/2,7/5, and 4/3. Which of the following can be used to evaluate the series

Step-by-step answer

P Answered by Specialist Expand and evaluate the series n=5 ai=2×i^2+3 (12) 24

D. Expand and evaluate the series n=5 ai=2×i^2+3 (13)

I JUST TOOK THE TEST, PICTURE ATTACHED.
Expand and evaluate the series n=5 ai=2×i^2+3 (14)

Mathematics

Evaluate the series 4-2 1-0.5 0.25 to s10. round to the nearest tenth. a. 5.50 b.-2.75 c.2.75 d. 2.66

Step-by-step answer

P Answered by Master Expand and evaluate the series n=5 ai=2×i^2+3 (15) 11

If we're talking about series, we're talking about a summation.
This is a geometric series, so we will be using the sum to n terms of a geometric series given by:

Expand and evaluate the series n=5 ai=2×i^2+3 (16), where a is the first term, r is the common ratio, and n is the number of terms.

Now, we know the first term is 4.
We know the ratio is Expand and evaluate the series n=5 ai=2×i^2+3 (17)
And we are finding to the 10th term.

Plugging everything into the equation:
Expand and evaluate the series n=5 ai=2×i^2+3 (18)

After this, we get: 2.6640625...≈ 2.66 (ie D)

Mathematics

Evaluate the series 4 – 2 + 1 – 0.5 + 0.25 to s10. round to the nearest hundredth.

Step-by-step answer

P Answered by PhD Expand and evaluate the series n=5 ai=2×i^2+3 (19) 6

The initial term of this geometric series is 4, and the common ratio is -1/2. The sum is given by

Expand and evaluate the series n=5 ai=2×i^2+3 (20)

The sum is approximately 2.66.

Mathematics

Evaluate the series 4 – 2 + 1 – 0.5 + 0.25 to s10. round to the nearest hundredth.

Step-by-step answer

P Answered by PhD Expand and evaluate the series n=5 ai=2×i^2+3 (21) 6

The initial term of this geometric series is 4, and the common ratio is -1/2. The sum is given by

Expand and evaluate the series n=5 ai=2×i^2+3 (22)

The sum is approximately 2.66.

Mathematics

We can use this power series to approximate the constant  . a) First, evaluate arctan(1) . (You do not need the series to evaluate it.) b) Use your answer from part (a) and the power series above to find a series representation for  . (The answer will be just a series – not a power series.) c) Verify that the series you found in part (b) converges. d) Use your convergent series from part (b) to approximate  with |error| 0.5. e) How many terms would you need to approximate  with |error| 0.001?

Step-by-step answer

P Answered by Specialist

(a)

Expand and evaluate the series n=5 ai=2×i^2+3 (23)

(b)

Expand and evaluate the series n=5 ai=2×i^2+3 (24)

(c)

Therefore if you sum any three terms of it you get the desired accuracy.

(d)

If you sum 1999 terms you get the desired accuracy.

Step-by-step explanation:

From the information given we know that

Expand and evaluate the series n=5 ai=2×i^2+3 (25)

(a)

For that you need to find and angle Expand and evaluate the series n=5 ai=2×i^2+3 (26) such that Expand and evaluate the series n=5 ai=2×i^2+3 (27)= 1, remember that

Expand and evaluate the series n=5 ai=2×i^2+3 (28) therefore

Expand and evaluate the series n=5 ai=2×i^2+3 (29)

(b)

Expand and evaluate the series n=5 ai=2×i^2+3 (30)

Then you just multiply by 4 and get that

Expand and evaluate the series n=5 ai=2×i^2+3 (31)

(c)

Using the alternating series test, since the sequence Expand and evaluate the series n=5 ai=2×i^2+3 (32) is decreasing and its limit tends to 0 when n tends to infinity the series is convergent.

(d)

Using the estimation theorem of alternating series we know that if Expand and evaluate the series n=5 ai=2×i^2+3 (33) denotes the partial sum of the series then

Expand and evaluate the series n=5 ai=2×i^2+3 (34)

Therefore we are looking for an Expand and evaluate the series n=5 ai=2×i^2+3 (35) such that

Expand and evaluate the series n=5 ai=2×i^2+3 (36)

we just have to solve that inequality, when you solve that inequality you get that

Expand and evaluate the series n=5 ai=2×i^2+3 (37)

Therefore if you sum any three terms of it you get the desired accuracy.

(e) For this part you need to solve the following inequality

Expand and evaluate the series n=5 ai=2×i^2+3 (38)

When you solve that inequality you get that

Expand and evaluate the series n=5 ai=2×i^2+3 (39)

so, if you sum 1999 terms you get the desired accuracy.

Mathematics

For the following telescoping series, find a formula for the nth term of the sequence of partial sums S n . Then evaluate lim n → [infinity] S n to obtain the value of the series or state that the series diverges. ∑ [infinity] k = 1 = 10 ( 5 k − 1 ) ( 5 k + 4 )

Step-by-step answer

P Answered by PhD

I'm guessing the sum is supposed to be

Expand and evaluate the series n=5 ai=2×i^2+3 (40)

Split the summand into partial fractions:

Expand and evaluate the series n=5 ai=2×i^2+3 (41)

Expand and evaluate the series n=5 ai=2×i^2+3 (42)

If Expand and evaluate the series n=5 ai=2×i^2+3 (43), then

Expand and evaluate the series n=5 ai=2×i^2+3 (44)

If Expand and evaluate the series n=5 ai=2×i^2+3 (45), then

Expand and evaluate the series n=5 ai=2×i^2+3 (46)

This means

Expand and evaluate the series n=5 ai=2×i^2+3 (47)

Consider the Expand and evaluate the series n=5 ai=2×i^2+3 (48)th partial sum of the series:

Expand and evaluate the series n=5 ai=2×i^2+3 (49)

The sum telescopes so that

Expand and evaluate the series n=5 ai=2×i^2+3 (50)

and as Expand and evaluate the series n=5 ai=2×i^2+3 (51), the second term vanishes and leaves us with

Expand and evaluate the series n=5 ai=2×i^2+3 (52)

Mathematics

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate Lim Sn to obtain the value of the series or state that the series diverges. n→[infinity] [infinity] Σ (4/√k+5 ) - 4/ √ k+6) k=1

Step-by-step answer

P Answered by PhD Expand and evaluate the series n=5 ai=2×i^2+3 (53) 1

Looks like the series is

Expand and evaluate the series n=5 ai=2×i^2+3 (54)

This series has n-th partial sum

Expand and evaluate the series n=5 ai=2×i^2+3 (55)

(where Expand and evaluate the series n=5 ai=2×i^2+3 (56) is used as a placeholder for the summand)

Expand and evaluate the series n=5 ai=2×i^2+3 (57)

In each grouped term, the last term is annihilated by the first term of the next group; that is, for instance,

Expand and evaluate the series n=5 ai=2×i^2+3 (58)

Ultimately, all the middle terms will vanish and we're left with

Expand and evaluate the series n=5 ai=2×i^2+3 (59)

As Expand and evaluate the series n=5 ai=2×i^2+3 (60), the last term converges to 0 and we're left with

Expand and evaluate the series n=5 ai=2×i^2+3 (61)

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