edexcel further math 2 help Watch

SennaRacer77
Badges: 0
Rep:
?
#1
Report Thread starter 7 years ago
#1
can someone explain to me how to use method of difference? like the one in page 12 of textbook ?
0
quote
reply
Potassium^2
Badges: 0
Rep:
?
#2
Report 7 years ago
#2
Basically write out the first few terms of the series until you spot a pattern for canceling.
Then follow this pattern and voila!
0
quote
reply
SennaRacer77
Badges: 0
Rep:
?
#3
Report Thread starter 7 years ago
#3
(Original post by Potassium^2)
Basically write out the first few terms of the series until you spot a pattern for canceling.
Then follow this pattern and voila!
may i know what is the pattern like? is it when they equal the same u cancel out? and something to do with diagonal?
0
quote
reply
dknt
Badges: 12
Rep:
?
#4
Report 7 years ago
#4
(Original post by SennaRacer77)
may i know what is the pattern like? is it when they equal the same u cancel out? and something to do with diagonal?
Consider the sum \displaystyle\sum_{r=1}^n \dfrac{1}{r} - \dfrac{1}{r+1}

The idea is to find terms which cancel and see which are left.. so we consider different vaules of r:

r=1: \ \ \ \ \ \ \ 1- \dfrac{1}{2}

r=2: \ \ \ \ \ \ \ \dfrac{1}{2}- \dfrac{1}{3}

r=3: \ \ \ \ \ \ \ \dfrac{1}{3}- \dfrac{1}{4}
...
r=n-1: \ \ \dfrac{1}{n-1}- \dfrac{1}{n}

r=n: \ \ \ \ \ \ \ \dfrac{1}{n}- \dfrac{1}{n+1}

You should be able to see that the half will cancel, the thirds will cancel. Also, although we haven't written it, the quarter will also cancel (if you imagine writing out the line for r=4)

As we continue to r=n, you should see that the 1/n will also cancel. Also, if you imagine writing out the line r=n-2, we would find that the 1/(n-1) would also cancel. Leaving us with only 2 terms. And so our summation becomes:

\displaystyle\sum_{r=1}^n \dfrac{1}{r} - \dfrac{1}{r+1} = 1 - \dfrac{1}{n+1} In other words, for this summation, whatever the value of n in the \displaystyle\sum_{r=1}^n \dfrac{1}{r} - \dfrac{1}{r+1} , the sum will equal 1 - \dfrac{1}{n+1}
1
quote
reply
SennaRacer77
Badges: 0
Rep:
?
#5
Report Thread starter 7 years ago
#5
(Original post by dknt)
Consider the sum \displaystyle\sum_{r=1}^n \dfrac{1}{r} - \dfrac{1}{r+1}

The idea is to find terms which cancel and see which are left.. so we consider different vaules of r:

r=1: \ \ \ \ \ \ \ 1- \dfrac{1}{2}

r=2: \ \ \ \ \ \ \ \dfrac{1}{2}- \dfrac{1}{3}

r=3: \ \ \ \ \ \ \ \dfrac{1}{3}- \dfrac{1}{4}
...
r=n-1: \ \ \dfrac{1}{n-1}- \dfrac{1}{n}

r=n: \ \ \ \ \ \ \ \dfrac{1}{n}- \dfrac{1}{n+1}

You should be able to see that the half will cancel, the thirds will cancel. Also, although we haven't written it, the quarter will also cancel (if you imagine writing out the line for r=4)

As we continue to r=n, you should see that the 1/n will also cancel. Also, if you imagine writing out the line r=n-2, we would find that the 1/(n-1) would also cancel. Leaving us with only 2 terms. And so our summation becomes:

\displaystyle\sum_{r=1}^n \dfrac{1}{r} - \dfrac{1}{r+1} = 1 - \dfrac{1}{n+1} In other words, for this summation, whatever the value of n in the \displaystyle\sum_{r=1}^n \dfrac{1}{r} - \dfrac{1}{r+1} , the sum will equal 1 - \dfrac{1}{n+1}
thanks man! i knida get the picture. if i need anything will let ya know
0
quote
reply
dknt
Badges: 12
Rep:
?
#6
Report 7 years ago
#6
(Original post by SennaRacer77)
thanks man! i knida get the picture. if i need anything will let ya know
No problem
0
quote
reply
SennaRacer77
Badges: 0
Rep:
?
#7
Report Thread starter 7 years ago
#7
btw,
im stuck at FP2 exercise 5E, page 99, i use my particular integral as lamda e^x and i get lamda=1, but i cant seem to get the answer :/

and exercise 5D page 97 i need help with that too :/
0
quote
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
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

  • University of Lincoln
    Mini Open Day at the Brayford Campus Undergraduate
    Wed, 19 Dec '18
  • University of East Anglia
    UEA Mini Open Day Undergraduate
    Fri, 4 Jan '19
  • Bournemouth University
    Undergraduate Mini Open Day Undergraduate
    Wed, 9 Jan '19

Were you ever put in isolation at school?

Yes (199)
27.95%
No (513)
72.05%

Watched Threads

View All