Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    11
    ReputationRep:
    The first three terms in the expansion of (a + 3x)^n are 16b, 144bx and 540bx^2.
    Find the values of the constants a, b and n.

    I'm pretty much clueless, and really need help! Thanks.
    Offline

    17
    ReputationRep:
    Well, what are the first 3 terms in the expansion of (a+3x)^n? Use the binomial theorem.
    Offline

    0
    ReputationRep:
    DFranklin strikes again with the most useless of posts.
    Offline

    0
    ReputationRep:
    I can get 'a' but Im stuck after that!
    • PS Helper
    Offline

    14
    PS Helper
    (Original post by Fraser_Integration)
    DFranklin strikes again with the most useless of posts.
    Lolwut? As far as I can tell his post was actually quite useful if you followed it.
    Offline

    2
    ReputationRep:
    (Original post by Fraser_Integration)
    {mod: abusive comment deleted}
    :lolwut:
    Fraser_Integration strikes again with the most useless of posts
    Offline

    0
    ReputationRep:
    (Original post by DFranklin)
    Well, what are the first 3 terms in the expansion of (a+3x)^n? Use the binomial theorem.
    That should give you 3 simultaneous equations.

    These sorts of questions always end up in simultaneous equations.
    Offline

    17
    ReputationRep:
    (Original post by stevencarrwork)
    That should give you 3 simultaneous equations.
    Sort of. They won't be terribly 'simultaneous' though. (i.e. you can easily solve them by substitution).

    These sorts of questions always end up in simultaneous equations.
    Not any more. Not that standards have fallen since our day, of course...
    Offline

    0
    ReputationRep:
    (Original post by DFranklin)
    Not any more. Not that standards have fallen since our day, of course...
    I always found those questions really dull. It was just processing symbols, with no thinking needed.

    You needed to memorise the binomial expansion, but once you did, the questions were so trivial.

    The questions always came down to '7b is 4 times (7*6/2))*b^2 What is b?'

    The questions were also never useful in other topics.

    How often do you have a binomial expansion where you know the second coefficient is 3 times the first, and the third is 8 times the second, and you have to work out what the power was?
    Offline

    0
    ReputationRep:
    This question appears on p.285 of OUP's Edexcel Core C1 C2 book. It's considerably harder than you'll meet in an actual exam, so don't be worried if you can't do it. Here's the solution for anyone still looking.

    First expand (a+3x)^n to the first three terms, applying the following binomial expansion formula:

    (a + bx)^n = (nC0)a^n + (nC1)a^(n-1)bx + (nC2)a^(n-2)(bx)^2 + ...

    If we do this correctly, we end up with the following three terms:

    (1) a^n
    (2) n * a^(n-1) * 3x = 3na^(n-1)x
    (3) n(n-1)/2! * a^(n-2) * (3x)^2 = 9/2(n)(n-1)a^(n-2)x^2

    We're told that these three terms correspond to 16b, 144bx, and 540bx^2 respectively. From this we obtain the following three simultaneous equations:

    (1) 16b = a^n
    (2) 144b = 3na^(n-1)
    (3) 540bx^2 = 9/2(n)(n-1)a^(n-2)

    So now we have to solve these for a, b, and n. But it's a little tricky, so I'm going to step through this part as well. Here goes. Looking at the first two equations, we see that 9 * 16b = 144b. So,

    9 * a^n = 3na^(n-1)
    3a^n = na^(n-1) [dividing both sides by 3]
    3/n = a^(n-1) / a^n [dividing both sides by n and a^n]
    3/n = a^-1 = 1/a
    [subtracting indices for division]
    a = n/3 [rearranging]
    n = 3a [rearranging]

    Looking at the last two equations, we see that 15/4 * 144b = 540b. So,

    15/4 * 3na^(n-1) = 9/2(n)(n-1)a^(n-2)
    5/2 * a^(n-1) = (n-1)a^(n-2) [dividing both sides by 9n/2]
    a^(n-1) / a^(n-2) = 2/5(n-1) [dividing both sides by 5/2 & a^(n-2)]
    a = 2/5(n-1) [subtracting indices for division]

    We now have two definitions of a: a = n/3 and a = 2/5(n-1). Therefore:

    n/3 = 2/5(n-1) [since a = a]
    5n = 6(n-1) = 6n-6 [multiplying both sides by 3*5]

    Now at last we're there!

    n = 6
    [simplifying above]
    a = 2 [since a = n/3]
    b = 4 [since 16b = a^n]
    Offline

    11
    ReputationRep:
    (Original post by Timotheos)
    Now at last we're there!
    At last is an understatement! 3 years and 22 days since the last post:rofl2:
 
 
 
  • 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.

  • Poll
    Would you like to hibernate through the winter months?
    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
  • 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.

  • 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.