Binomial Theorem


    Rep:
    Hi need some help with a binomial theorem question.

    Evaluate (1-sqrt2)^5. Give your answer in terms of sqrt2

    Using pascals triangle I get

    (x^5)+(5(x^4)y)+(10(x^3)(y^2)+(1 0(x^2)(y^3))+(5x(y^4))+(y^5)

    After substituting in the x and y I get

    31-(5*sqrt2)-(10*sqrt2^2)-(10*sqrt2^3)-(5*sqrt2^4)-(sqrt2^5)

    But im pretty sure this isnt right?
    Any clues to where im going wrong?

    Rep:
    I assume you mean

    (x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5

    Then for

    (1-\sqrt{x})^5

    x = 1

    and

    y = -\sqrt{x}

    so you need to check your substitution

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: April 22, 2012

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