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Thread: Binomial Theorem HSC 1985 Q6

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    Binomial Theorem HSC 1985 Q6

    Could anyone help me with part ii of this question?

    I finished part i and got 14. Im guessing part ii has something to do with using the 3+2x and maybe n=10 and some sort of sub for x. Part from that I'm lost.

    Any help appreciated, thanks.

    Screen_Shot_2018_06_17_at_4_18_55_pm.png

    Screen_Shot_2018_06_17_at_4_19_04_pm.png

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    Re: Binomial Theorem HSC 1985 Q6



    Then letting x = 1 and then x = -1, you get:



    Adding the 2, the terms with opposing signs cancel out, leaving:



    Simplifying, you get the required result.




    ps: my stoopid internet connection keeps dropping off! Hence taking so long to post above solution.
    Last edited by Drongoski; 17 Jun 2018 at 11:02 PM.
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    Re: Binomial Theorem HSC 1985 Q6

    Quote Originally Posted by boredofstudiesuser1 View Post
    Could anyone help me with part ii of this question?

    I finished part i and got 14. Im guessing part ii has something to do with using the 3+2x and maybe n=10 and some sort of sub for x. Part from that I'm lost.

    Any help appreciated, thanks.

    Screen_Shot_2018_06_17_at_4_18_55_pm.png

    Screen_Shot_2018_06_17_at_4_19_04_pm.png
    Pretty much bang on there. The other hint I would give is notice that you only see every second binomial coefficient. The first step to derive such a series is to substitute a value for x and then substitute the negative of that value.

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    Re: Binomial Theorem HSC 1985 Q6

    Quote Originally Posted by Drongoski View Post
    Edit: Great, thank you so much.
    Last edited by boredofstudiesuser1; 17 Jun 2018 at 6:38 PM.

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    Re: Binomial Theorem HSC 1985 Q6

    Quote Originally Posted by Trebla View Post
    Pretty much bang on there. The other hint I would give is notice that you only see every second binomial coefficient. The first step to derive such a series is to substitute a value for x and then substitute the negative of that value.
    Thank you.

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