# A Level Further Maths: Series QuestionsWatch

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Thread starter 1 month ago
#1
Just can't seem to work out the answers to these questions, hope someone here can help

1) a. By considering ∑((𝑟 + 1)^5 − 𝑟^5), show that
n
∑ 𝑟^4 = 1/30𝑛(𝑛 + 1)(2𝑛 + 1)(3𝑛^2 + 3𝑛 − 1)
r=1

1) b. Hence find the value of 50^4 + 51^4 + ⋯+ 80^4

2) a. Show that (𝑟 + 1)! − (𝑟 − 1)! = (𝑟2 + 𝑟 − 1)(𝑟 − 1)!

2) b. Hence show that
n
∑ (𝑟^2 + 𝑟 − 1)(𝑟 − 1)! = (𝑛 + 2)𝑛! − 2
r=1

3) a. Show that
(2 𝑟)/((𝑟 + 1)(𝑟 + 2)) = 1/𝑟 − 2/( 𝑟 + 1) + 1/( 𝑟 + 2)

3) b. Hence determine an expression for
n
∑ 1/(𝑟(𝑟 + 1)(𝑟 + 2))
r=1
1/4 − 1/2 * f(𝑛)

3) c. Find the value of

∑ 1/(𝑟(𝑟 + 1)(𝑟 + 2))
r=1
0
1 month ago
#2
(Original post by hammerthrow)
Just can't seem to work out the answers to these questions, hope someone here can help

1) a. By considering ∑((𝑟 + 1)^5 − 𝑟^5), show that
n
∑ 𝑟^4 = 1/30𝑛(𝑛 + 1)(2𝑛 + 1)(3𝑛^2 + 3𝑛 − 1)
r=1

1) b. Hence find the value of 50^4 + 51^4 + ⋯+ 80^4

Well, tackle them one by one. For the first one, you can use Method of Differences to state what

is in terms of . Let's call this result because I cannot be bothered to work it out and I'm leaving it to you intentionally.

Once you have that, a realisation is required to notice that

Therefore;

And you can treat the RHS as .

You can determine all of these sums except the one. Once you do that, just rearrange the equality

for
0
Thread starter 1 month ago
#3
(Original post by RDKGames)
Well, tackle them one by one. For the first one, you can use Method of Differences to state what

is in terms of . Let's call this result because I cannot be bothered to work it out and I'm leaving it to you intentionally.

Once you have that, a realisation is required to notice that

Therefore;

And you can treat the RHS as .

You can determine all of these sums except the one. Once you do that, just rearrange the equality

for
Any clue how to approach the questions 3 a b and c
0
1 month ago
#4
(Original post by hammerthrow)
Any clue how to approach the questions 3 a b and c
Have you ever covered Partial Fraction Decomposition ?
0
Thread starter 1 month ago
#5
(Original post by RDKGames)
Have you ever covered Partial Fraction Decomposition ?
not at all
0
1 month ago
#6
(Original post by hammerthrow)
not at all
Then you can't do part (a).
0
Thread starter 1 month ago
#7
(Original post by RDKGames)
Then you can't do it.
Wow my teacher is very clever 🙄
0
1 month ago
#8
(Original post by hammerthrow)
Wow my teacher is very clever 🙄
If you have covered Method of Differences (or telescoping sums) you can do parts (b) and (c) by just taking the result of part (a) at face value.
0
Thread starter 1 month ago
#9
(Original post by RDKGames)
If you have covered Method of Difference (or telescoping sums) you can do parts (b) and (c) by just taking the result of part (a) at face value.
yeah we have covered method of difference. surely part c is just infinity because u can't add numbers to it ect
0
1 month ago
#10
(Original post by hammerthrow)
yeah we have covered method of difference. surely part c is just infinity because u can't add numbers to it ect
Not quite. If you work out the answer to part (b) correctly then you would notice that is a rational fraction with a linear term in the numerator, and a quadratic term in the denominator.

The upper limit of can be rewritten as a limit; i.e. .

Just like improper integrals are written in terms of limits (if you have covered those). And clearly you will need to be taking the limit of . Key point here is that in this limit precisely because of what type of function it is.

So, f(n) disappears and the sum at infinity is just a finite value, whatever is leftover.
0
Thread starter 1 month ago
#11
(Original post by RDKGames)
Not quite. If you work out the answer to part (b) correctly then you would notice that is a rational fraction with a linear term in the numerator, and a quadratic term in the denominator.

The upper limit of can be rewritten as a limit; i.e. .

Just like improper integrals are written in terms of limits (if you have covered those). And clearly you will need to be taking the limit of . Key point here is that in this limit precisely because of what type of function it is.

So, f(n) disappears and the sum at infinity is just a finite value, whatever is leftover.
Oh right okay. would have never thought of it like that tbh
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