A Level Further Maths: Series Questions Watch

hammerthrow
Badges: 6
Rep:
?
#1
Report 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
giving your answer in the form
1/4 − 1/2 * f(𝑛)

3) c. Find the value of

∑ 1/(𝑟(𝑟 + 1)(𝑟 + 2))
r=1
0
reply
RDKGames
Badges: 20
Rep:
?
#2
Report 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

\displaystyle \sum_{r=1}^n (r+1)^5 - r^5

is in terms of n. Let's call this result f(n) 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 (r+1)^5 - r^5 = 5r^4 + 10r^3 + 10r^2 + 5r + 1

Therefore;

\displaystyle \sum_{r=1}^n (r+1)^5 - r^5 = f(n) = \sum_{r=1}^n 5r^4 + 10r^3 + 10r^2 + 5r + 1

And you can treat the RHS as \displaystyle 5\sum_{r=1}^n r^4 + 10\sum_{r=1}^n r^3 + 10\sum_{r=1}^nr^2 + 5\sum_{r=1}^n r + \sum_{r=1}^n 1.

You can determine all of these sums except the r^4 one. Once you do that, just rearrange the equality

f(n) =\displaystyle  5\sum_{r=1}^n r^4 + 10\sum_{r=1}^n r^3 + 10\sum_{r=1}^nr^2 + 5\sum_{r=1}^n r + \sum_{r=1}^n 1

for  \displaystyle \sum_{r=1}^n r^4
0
reply
hammerthrow
Badges: 6
Rep:
?
#3
Report 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

\displaystyle \sum_{r=1}^n (r+1)^5 - r^5

is in terms of n. Let's call this result f(n) 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 (r+1)^5 - r^5 = 5r^4 + 10r^3 + 10r^2 + 5r + 1

Therefore;

\displaystyle \sum_{r=1}^n (r+1)^5 - r^5 = f(n) = \sum_{r=1}^n 5r^4 + 10r^3 + 10r^2 + 5r + 1

And you can treat the RHS as \displaystyle 5\sum_{r=1}^n r^4 + 10\sum_{r=1}^n r^3 + 10\sum_{r=1}^nr^2 + 5\sum_{r=1}^n r + \sum_{r=1}^n 1.

You can determine all of these sums except the r^4 one. Once you do that, just rearrange the equality

f(n) =\displaystyle  5\sum_{r=1}^n r^4 + 10\sum_{r=1}^n r^3 + 10\sum_{r=1}^nr^2 + 5\sum_{r=1}^n r + \sum_{r=1}^n 1

for  \displaystyle \sum_{r=1}^n r^4
Any clue how to approach the questions 3 a b and c
0
reply
RDKGames
Badges: 20
Rep:
?
#4
Report 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
reply
hammerthrow
Badges: 6
Rep:
?
#5
Report Thread starter 1 month ago
#5
(Original post by RDKGames)
Have you ever covered Partial Fraction Decomposition ?
not at all
0
reply
RDKGames
Badges: 20
Rep:
?
#6
Report 1 month ago
#6
(Original post by hammerthrow)
not at all
Then you can't do part (a).
0
reply
hammerthrow
Badges: 6
Rep:
?
#7
Report Thread starter 1 month ago
#7
(Original post by RDKGames)
Then you can't do it.
Wow my teacher is very clever 🙄
0
reply
RDKGames
Badges: 20
Rep:
?
#8
Report 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
reply
hammerthrow
Badges: 6
Rep:
?
#9
Report 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
reply
RDKGames
Badges: 20
Rep:
?
#10
Report 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 f(n) is a rational fraction with a linear term in the numerator, and a quadratic term in the denominator.

The upper limit of \infty can be rewritten as a limit; i.e. \displaystyle \sum_{r=1}^{\infty} = \lim_{n\to\infty} \sum_{r=1}^n.

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 f(n). Key point here is that f(n) \to 0 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
reply
hammerthrow
Badges: 6
Rep:
?
#11
Report 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 f(n) is a rational fraction with a linear term in the numerator, and a quadratic term in the denominator.

The upper limit of \infty can be rewritten as a limit; i.e. \displaystyle \sum_{r=1}^{\infty} = \lim_{n\to\infty} \sum_{r=1}^n.

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 f(n). Key point here is that f(n) \to 0 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
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • SOAS University of London
    Development Studies, Interdisciplinary Studies, Anthropology and Sociology, Languages, Cultures and Linguistics, Arts, Economics, Law, History, Religions and Philosophies, Politics and International Studies, Finance and Management, East Asian Languages & Cultures Postgraduate
    Sat, 25 Jan '20
  • University of Huddersfield
    Undergraduate Open Day Undergraduate
    Sat, 25 Jan '20
  • The University of Law
    Solicitor Series: Assessing Trainee Skills – LPC, GDL and MA Law - Guildford campus Postgraduate
    Wed, 29 Jan '20

How many uni open days have you been to/did you go to?

0 (67)
27.57%
1 (40)
16.46%
2 (40)
16.46%
3 (34)
13.99%
4 (16)
6.58%
5 (20)
8.23%
6 (6)
2.47%
7+ (20)
8.23%

Watched Threads

View All
Latest
My Feed