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

    0
    ReputationRep:
    The coefficient of x2
    in the binomial expansion of (1 + 0.4x)n
    , where n is a positive integer, is 1.6

    a Find the value of n.
    b Use your value of n to find the coefficient of x
    4
    in the expansion.

    need help on a, have no idea where to start,
    thanks =)
    Offline

    3
    ReputationRep:
    Firstly, do you know how to work out the binomial expansion?
    • Community Assistant
    Offline

    19
    ReputationRep:
    Expand (1+0.4x)^n as far as the term in x^2

    Set the coefficient of x^2 equal to 1.6.
    Offline

    13
    ReputationRep:
    Compare coefficient of the x^2 term
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Noble.)
    Firstly, do you know how to work out the binomial expansion?
    yes, using NcR notation.
    Offline

    3
    ReputationRep:
    (Original post by Hi, How are you ?)
    yes, using NcR notation.
    Ok, Mr. M has said how to proceed once you've done the binomial expansion up to the x^2 term.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mr M)
    Expand (1+0.4x)^n as far as the term in x^2

    Set the coefficient of x^2 equal to 1.6.
    Tried using NcR notation, and then I got
    \displaystyle \binom{n}{0} ^ 1n +\displaystyle \binom{n}{1} ^ 1n-1 ^ 0.4x + \displaystyle \binom{n}{2} ^ 1n-2 ^ (0.4x)2
    , What do i do next ?
    Offline

    3
    ReputationRep:
    (1+x)^n = {n \choose 0}x^0 + {n \choose 1}x^1 + {n \choose 2}x^2 +  \cdots + {n \choose {n-1}}x^{n-1} + {n \choose n}x^n

    So

    (1+0.4x)^n = {n \choose 0}(0.4x)^0 + {n \choose 1}(0.4x)^1 + {n \choose 2}(0.4x)^2 +  \cdots

    You know that {n \choose 2}(0.4x)^2 = 1.6x^2

    You can use this to determine n.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Noble.)
    (1+x)^n = {n \choose 0}x^0 + {n \choose 1}x^1 + {n \choose 2}x^2 +  \cdots + {n \choose {n-1}}x^{n-1} + {n \choose n}x^n

    So

    (1+0.4x)^n = {n \choose 0}(0.4x)^0 + {n \choose 1}(0.4x)^1 + {n \choose 2}(0.4x)^2 +  \cdots

    You know that {n \choose 2}(0.4x)^2 = 1.6x^2

    You can use this to determine n.
    How would you get the n out of the brackets, what process would you do??
    • Community Assistant
    Offline

    19
    ReputationRep:
    (Original post by Hi, How are you ?)
    How would you get the n out of the brackets, what process would you do??
    \displaystyle \binom{n}{r} = \frac{n!}{r!(n-r)!}

    but the easiest way is just to write down a few rows of Pascal's triangle!
    Offline

    3
    ReputationRep:
    (Original post by Hi, How are you ?)
    How would you get the n out of the brackets, what process would you do??
    By definition  {n \choose m} = \dfrac{n!}{m!(n-m)!}
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Noble.)
    By definition  {n \choose m} = \dfrac{n!}{m!(n-m)!}
    so it means that \displaystyle \binom{n}{2}

    = n! / 2!(n-2)!
    = n! / 2(n-2)!
    Offline

    3
    ReputationRep:
    (Original post by Hi, How are you ?)
    so it means that \displaystyle \binom{n}{2}

    = n! / 2!(n-2)!
    = n! / 2(n-2)!
    Yep. Now note that n! = (n-1)(n-2)(n-3) x ... x 1 and (n-2)! = (n-2)(n-3)(n-4) x... x 1

    So you should be able to see what terms cancel in the fraction.
    • Community Assistant
    Offline

    19
    ReputationRep:
    (Original post by Noble.)
    Yep. Now note that n! = (n-1)(n-2)(n-3) x ... x 1 and (n-2)! = (n-2)(n-3)(n-4) x... x 1

    So you should be able to see what terms cancel in the fraction.
    Flippin' heck. Isn't this all a bit OTT?

    1

    1 1

    1 2 1

    1 3 3 1

    1 4 6 4 1

    1 5 10 10 5 1
    Offline

    3
    ReputationRep:
    (Original post by Mr M)
    Flippin' heck. Isn't this all a bit OTT?

    1

    1 1

    1 2 1

    1 3 3 1

    1 4 6 4 1

    1 5 10 10 5 1
    Haha, you're right - although I do hate Pascal's triangle

    Once you get familiarised with nCr though it obviously isn't anywhere near as cumbersome and you can just write

     {n \choose 2} = \dfrac{n(n-1)}{2}
 
 
 
Poll
If you won £30,000, which of these would you spend it on?
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.