Turn on thread page Beta
    • Thread Starter
    Offline

    9
    ReputationRep:
    Here was my attempt at the question, using a=49 and d=-3 which this examcafe solution uses. However, they use sn=n/2[a+l] whereas I used Sn=n/2[2a+(n-1)d].

    The question was this: find the value of the sum of the first 15 terms of the series.
    I got part A correct, but not part B.

    Here was my solution: Name:  IMG_20170115_155334.jpg
Views: 146
Size:  495.7 KB

    Where did I go wrong?

    I have attached the examcafe solution to this.
    Attachment 611766611768
    Attached Images
     
    Offline

    9
    ReputationRep:
    Name:  16121568_1175253342529675_1897466642_o.jpg
Views: 126
Size:  96.1 KB
    Offline

    1
    ReputationRep:
    (Original post by blobbybill)
    Here was my attempt at the question, using a=49 and d=-3 which this examcafe solution uses. However, they use sn=n/2[a+l] whereas I used Sn=n/2[2a+(n-1)d].

    The question was this: find the value of the sum of the first 15 terms of the series.
    I got part A correct, but not part B.

    Here was my solution: Name:  IMG_20170115_155334.jpg
Views: 146
Size:  495.7 KB

    Where did I go wrong?

    I have attached the examcafe solution to this.
    Attachment 611766611768
    You went from  \frac{15}{2} (56) to  15 \times 112
    Try that again. It should give you  15 \times 28 because  56 \div 2 = 28
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by rashid.mubasher)
    You went from  \frac{15}{2} (56) to  15 \times 112
    Try that again. It should give you  15 \times 28 because  56 \div 2 = 28
    How would I know whether I was able to do it the way I did, or whether I had to do it like you just said. Also, how od you know that you can make the 15/2 into 15 by halving the 56? I don't understand that.

    And why was it wrong for me to multiply it by 2 in order to eliminate the fraction. In maths before, I have been able to multiply something by the denominator in order to make it easier to solve as you have eliminated the fraction. I don't get it at all. Is that only possible when you are simplifying something, not solving?
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by wightsnowolf)
    Name:  16121568_1175253342529675_1897466642_o.jpg
Views: 126
Size:  96.1 KB
    How do you get from 15/2(98-42) to 420 so quick? How do you know that you can get rid of the 15/2 and make it into 15 by halving the stuff inside the bracket?

    And why was it wrong for me to multiply it by 2 in order to eliminate the fraction. In maths before, I have been able to multiply something by the denominator in order to make it easier to solve as you have eliminated the fraction. I don't get it at all. Is that only possible when you are simplifying something, not solving?
    Offline

    9
    ReputationRep:
    (Original post by blobbybill)
    How do you get from 15/2(98-42) to 420 so quick? How do you know that you can get rid of the 15/2 and make it into 15 by halving the stuff inside the bracket?

    And why was it wrong for me to multiply it by 2 in order to eliminate the fraction. In maths before, I have been able to multiply something by the denominator in order to make it easier to solve as you have eliminated the fraction. I don't get it at all. Is that only possible when you are simplifying something, not solving?
    By multiplying by two you effectively began to work out what 2 times the sum of the first 15 terms was, however in your working you actually multiplied by 4 since you multiplied both the inside and the outside of the brackets by 2.

    You should have just calculated the value as it was, there's no need to try and change it.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by wightsnowolf)
    By multiplying by two you effectively began to work out what 2 times the sum of the first 15 terms was, however in your working you actually multiplied by 4 since you multiplied both the inside and the outside of the brackets by 2.

    You should have just calculated the value as it was, there's no need to try and change it.
    How could I have calculated the value as it was?

    I didn't know how to calculate 15/2(56). How would you do that in your head? That's why I multiplied it (i didnt know it was wrong at the time). So how would you calculate it? Some people are saying you can just do 15(56/2) to give you 15*28. Why is that possible to do? Why are you able to change the 15/2 into 15 and change the 56 into 56/2?

    Why doesn't dividing a different part of it by 2 instead of the original 15 affect the answer?
    Offline

    9
    ReputationRep:
    (Original post by blobbybill)
    How could I have calculated the value as it was?

    I didn't know how to calculate 15/2(56). How would you do that in your head? That's why I multiplied it (i didnt know it was wrong at the time). So how would you calculate it? Some people are saying you can just do 15(56/2) to give you 15*28. Why is that possible to do? Why are you able to change the 15/2 into 15 and change the 56 into 56/2?

    Why doesn't dividing a different part of it by 2 instead of the original 15 affect the answer?
    Multiplication is associative so (15/2)(56) is the same as (15)(56/2).

    So you'd calculate 15(56/2) = 15(28) = 420.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by wightsnowolf)
    Multiplication is associative so (15/2)(56) is the same as (15)(56/2).

    So you'd calculate 15(56/2) = 15(28) = 420.
    Ah, that explains it, thanks.

    I definitely remember in C1 being able to multiply something by the denominator in order to eliminate a fraction. I think I did it in C1 coordinate geometry line equations, eg when I had y=mx+c or y-y1=m(x-x1) or ax + by + c=0. WHy could i do that in that scenario, whereas in this scenario, I am not able to multiply it by the numerator in order to eliminate the fraction?

    In future, how would I know whether I am able to multiply my expressions by the denominator (to eliminate the fractions), or whether I am not able to do that (like in this case)? I am confused about it.

    Thanks a lot!
    Offline

    9
    ReputationRep:
    (Original post by blobbybill)
    Ah, that explains it, thanks.

    I definitely remember in C1 being able to multiply something by the denominator in order to eliminate a fraction. I think I did it in C1 coordinate geometry line equations, eg when I had y=mx+c or y-y1=m(x-x1) or ax + by + c=0. WHy could i do that in that scenario, whereas in this scenario, I am not able to multiply it by the numerator in order to eliminate the fraction?

    In future, how would I know whether I am able to multiply my expressions by the denominator (to eliminate the fractions), or whether I am not able to do that (like in this case)? I am confused about it.

    Thanks a lot!
    Think about it this way. Whatever you do to one side of the equation you have to do to the other. So in the case of this question, when you tried to multiply everything by two to cancel out the fraction, the other side of the equation should have also been multiplied. I'll attach a picture to show what you tried to do, which should show you where the mistake came from.

    As long as you make sure that you're doing the same thing to both sides then you'll not run into problems like this.

    Name:  16111492_1175522382502771_1072656539_n.jpg
Views: 115
Size:  31.7 KB
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 15, 2017
Poll
Do you think parents should charge rent?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.