Join TSR now and get all your revision questions answeredSign up now
    • Thread Starter
    Offline

    2
    ReputationRep:
    I am stuck on this question:
    given that (1+x)^n = 1 + 12x + x^2 find the value of a

    Does n= 2 ? Im not sure on how to do this question
    Offline

    0
    ReputationRep:
    (Original post by Alex-Torres)
    I am stuck on this question:
    given that (1+x)^n = 1 + 12x + x^2 find the value of a

    Does n= 2 ? Im not sure on how to do this question
    Where's a being defined? all I can see are n's and x's, unless you mean (1+ax) but still then its a bit cryptic. And no, n does not equal 2 since using the laws of expansion
    (1+x)^n=1^n+nx+....
    Offline

    2
    ReputationRep:
    (Original post by Alex-Torres)
    I am stuck on this question:
    given that (1+x)^n = 1 + 12x + x^2 find the value of a

    Does n= 2 ? Im not sure on how to do this question
    There isn't an 'a' in the question?
    Offline

    15
    ReputationRep:
    was it asking for

    ( 1+ax )n ?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Robbie242)
    Where's a being defined? all I can see are n's and x's, unless you mean (1+ax) but still then its a bit cryptic. And no, n does not equal 2 since using the laws of expansion
    (1+x)^n=1^n+nx+....
    (Original post by Felix Felicis)
    There isn't an 'a' in the question?
    (Original post by the bear)
    was it asking for

    ( 1+ax )n ?
    Sorry. Given that (1+x)^n = 1 + 12x + ax^2, find the value of a.
    Offline

    0
    ReputationRep:
    (Original post by Alex-Torres)
    Sorry. Given that (1+x)^n = 1 + 12x + ax^2, find the value of a.
    Notice that from earlier the 2nd term 12x=nx divide by x, and we have n=12, I'm sure you know how to solve for a now

    Spoiler:
    Show
    Find the x^2 term in the expansion and set it equal to ax^2
    Offline

    15
    ReputationRep:
    well the x2 term is given by

    {n( n-1 )/2! }x2
    Offline

    2
    ReputationRep:
    (Original post by Alex-Torres)
    Sorry. Given that (1+x)^n = 1 + 12x + ax^2, find the value of a.
    Should this be (1+x)^{n} = 1 + 12x + ax^{2} + ... or does it stop at the x^2 term?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Felix Felicis)
    Should this be (1+x)^{n} = 1 + 12x + ax^{2} + ... or does it stop at the x^2 term?
    It stops at the ax^2
    Offline

    0
    ReputationRep:
    (Original post by Alex-Torres)
    It stops at the ax^2
    That can't be right? By the information you've given me I've deduced that n=12, it can't be untrue as its part of the standard expansion, I think it goes on for longer otherwise people would just say n=2 herp derp which isn't true for the 2nd term
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Robbie242)
    Notice that from earlier the 2nd term 12x=nx divide by x, and we have n=12, I'm sure you know how to solve for a now

    Spoiler:
    Show
    Find the x^2 term in the expansion and set it equal to ax^2
    I got 66?
    Attached Images
     
    Offline

    0
    ReputationRep:
    (Original post by Alex-Torres)
    I got 66?
    That is correct, now set it equal to ax^2 and solve for a also a quicker way for terms like this \frac{n(n-1)}{2!}(x^2) if in the form (1+x)^n
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Robbie242)
    That is correct, now set it equal to ax^2 and solve for a also a quicker way for terms like this \frac{n(n-1)}{2!}(x^2) if in the form (1+x)^n
    Doesnt a just equal 66 and thats the end of the question?
    Offline

    0
    ReputationRep:
    (Original post by Alex-Torres)
    Doesnt a just equal 66 and thats the end of the question?
    and that is indeed, try not to forget if you expand up to a term to leave +.... some examiners may be very picky about this
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Robbie242)
    and that is indeed, try not to forget if you expand up to a term to leave +.... some examiners may be very picky about this
    Thanks!
    Do you know where I went wrong in this question?Name:  image.jpg
Views: 78
Size:  142.2 KB
    Offline

    0
    ReputationRep:
    (Original post by Alex-Torres)
    Thanks!
    Do you know where I went wrong in this question?Name:  image.jpg
Views: 78
Size:  142.2 KB
    Try be more careful in your workings and re-expand, one of the terms I got was to be -45, just re-expand more carefully
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Robbie242)
    Try be more careful in your workings and re-expand, one of the terms I got was to be -45, just re-expand more carefully
    Could you please take a picture of your working, as I can't see how you got that answer?
    Offline

    0
    ReputationRep:
    (Original post by Alex-Torres)
    Could you please take a picture of your working, as I can't see how you got that answer?
    Sure after dinner if your still stuck, will be like 10 mins
    Offline

    1
    ReputationRep:
    (Original post by Alex-Torres)
    Thanks!
    Do you know where I went wrong in this question?Name:  image.jpg
Views: 78
Size:  142.2 KB
    Man, root3^5 is root3 x root3 x root3 x root3 x root 3... Which gives root 243 which is 9root 3.
    And similarly for the rest!
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Robbie242)
    Sure after dinner if your still stuck, will be like 10 mins
    Im OK now, but thanks anyway! And the above poster
 
 
 
Poll
Which Fantasy Franchise is the best?
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

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