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Thread: Using Binomial Theorem

  1. #1
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    Using Binomial Theorem

    If n is a positive integer, prove that the integral part of [IMG]file:///C:/Users/user/AppData/Local/Temp/msohtmlclip1/01/clip_image002.png[/IMG] is an even integer.
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    Exclamation Re: Using Binomial Theorem

    If n is a positive integer, prove that the integral part of (5√5+11)^(2n+1) is an even no.

    I tried a lot on this ques but ans coming to be odd, which is incorrect.
    I doubt my soln. would be of any help to u all.

    Sorry for the first post.It is very difficult to write equations and symbols in this forum.
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  3. #3
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    Re: Using Binomial Theorem

    I would use induction as the overall strategy.

    In the course of proving $P_n \Rightarrow P_{n+1}$ you're going to have to apply the binomial theorem.

    See where that leads you.
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  4. #4
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    Re: Using Binomial Theorem

    using the binomial theorem it is easy to see that there are integers $a_n$ and $b_n$ such that

    $\displaystyle x_n=\left(5\sqrt{5}+11\right)^{2n+1}=a_n+b_n\sqrt{ 5}$

    and

    $\displaystyle y_n=\left(5\sqrt{5}-11\right)^{2n+1}=-a_n+b_n\sqrt{5}$

    therefore

    $x_n=2a_n+y_n$ with $0<y_n<1$

    so

    $\left\lfloor x_n\right\rfloor =2a_n$
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  5. #5
    Forum Admin topsquark's Avatar
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    Re: Using Binomial Theorem

    Quote Originally Posted by mak29 View Post
    Sorry for the first post.It is very difficult to write equations and symbols in this forum.
    We have a LaTeX system here. It's pretty easy to learn the basics, like what you were trying to post.

    -Dan
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  6. #6
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    Re: Using Binomial Theorem

    What is wrong in my solution:

    〖(5√5+11)〗^(2n+1)=I + f ; where I is the integral part and f is a fraction;

    〖(5√5-11)〗^(2n+1)=f1; where f1 is a pure fraction .

    0<f<1
    0<f1<1

    On adding:
    I+f+f1=2(2n+1C0(〖(5√5)^2n+1 ).......)
    ,i.e. I+f+f1=2k where k is a constant

    0<f+f1<2
    .i.e. f+f1=1


    I=2k-1 ; which is a odd no.
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  7. #7
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    Re: Using Binomial Theorem

    Quote Originally Posted by mak29 View Post
    What is wrong in my solution:

    〖(5√5+11)〗^(2n+1)=I + f ; where I is the integral part and f is a fraction;

    〖(5√5-11)〗^(2n+1)=f1; where f1 is a pure fraction .

    0<f<1
    0<f1<1

    On adding:
    I+f+f1=2(2n+1C0(〖(5√5)^2n+1 ).......)
    ,i.e. I+f+f1=2k where k is a constant

    0<f+f1<2
    .i.e. f+f1=1


    I=2k-1 ; which is a odd no.
    $\displaystyle I + f + f_1 = 2k$

    Here $k$ is not an integer

    Also,

    $f + f_1$ is not an integer and it does not equal $1$
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