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higher maths

  • 14-03-2010 6:39pm
    #1
    ted9308
    Registered Users, Registered Users 2 Posts: 82 ✭✭


    Could anyone help me with this question just cant seem to get it out right-binomial theorem by the way "if h is a constant and hx(to the power of 4)y(squared) is a term of the expansion of (3x -4ysquared)to the power of n..find the value of n and h:confused::confused:


Comments

  • #2
    straight_As
    Closed Accounts Posts: 193 ✭✭straight_As


    Post up what you've done so far and boards might be able to point you in the right direction.

    It's pointless someone just posting up a solution though. :)


  • #3
    [Deleted User]
    Posts: 0 ✭ [Deleted User]


    Term with: hx^4y^2 of BE: (3x-4y^2)^n

    General term: Ur+1= nCr (3x)^n-r (-4y^2)^r

    =nCr 3^n-r . x^n-r. (-4)^r. y^2r
    = nCr (3^n-r. [-4]^r) x^n-r. y^2r

    Power of x has to =4, and power of y has to =2
    n-r=4
    2r=2 =) r=1
    n-1=4
    n=5

    U1+1= 5C1 (3^5-1. -4^1) x^5-1 y^2r
    U2= 5C1 (3^4.-4) x^4 y^2

    h= coefficient of term= 5C1 (81.-4) = 5 . -324 = -1620

    U2= -1620x^4y^2

    n=5, h= -1620




    I think this is right, do you have the answer to check it with? Sorry if this is hard to follow! (^ means to the power of btw)


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