Results 1 to 8 of 8

Thread: Binomial expansion

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Binomial expansion

    Find the coefficients of $\displaystyle x^7$ in the expansion of $\displaystyle (1-3x^2)(x^2-\frac{2}{x})^7$

    My attempt:

    to obtain $\displaystyle x^7$, coefficients of $\displaystyle x^5$ and $\displaystyle x^7$ have to be used.

    $\displaystyle T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r$

    therefore, $\displaystyle 2(7-r)+(-r)=5$
    $\displaystyle r=3$


    OR $\displaystyle 2(7-r)+(-r)=5$
    $\displaystyle r= 2(\frac{1}{3})$

    Puzzled at arriving at $\displaystyle 2\frac{1}{3}$ for $\displaystyle r$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Thanks
    1
    Quote Originally Posted by Punch View Post
    Find the coefficients of $\displaystyle x^7$ in the expansion of $\displaystyle (1-3x^2)(x^2-\frac{2}{x})^7$

    My attempt:

    to obtain $\displaystyle x^7$, coefficients of $\displaystyle x^5$ and $\displaystyle x^7$ have to be used.

    $\displaystyle T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r$

    therefore, $\displaystyle 2(7-r)+(-r)=5$
    $\displaystyle r=3$


    OR $\displaystyle 2(7-r)+(-r)=5$
    $\displaystyle r= 2(\frac{1}{3})$

    Puzzled at arriving at $\displaystyle 2\frac{1}{3}$ for $\displaystyle r$.
    is the answer 840 ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by mathaddict View Post
    is the answer 840 ?
    yes it is! please show me the right way of solving it
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Thanks
    1
    Quote Originally Posted by Punch View Post
    yes it is! please show me the right way of solving it
    Great . Of course , we can expand the whole thing and combine terms which will produce x^7 but it would be tedious and risky so its better to use the binomial theorem :

    $\displaystyle
    (a+b)^n=\sum^{n}_{k=0} \binom{n}{k} a^{n-k}b^k
    $

    $\displaystyle
    (x^2-\frac{2}{x})^7=\sum^{n}_{k=0} \binom{7}{k}x^{14-3k}\cdot 2^k
    $

    $\displaystyle (1-3x^2)(x^2-\frac{2}{x})^7$

    $\displaystyle =\sum^{n}_{k=0} \binom{7}{k}x^{14-3k}\cdot 2^k+\sum^{n}_{k=0}\binom{7}{k} x^{16-3k}\cdot 3\cdot 2^k$

    so now lets see what values of k give 7

    14-3k=7 , k=7/3 and this cant be since k must be +ve integer

    how about 16-3k=7 , k=3 .

    its coefficient is $\displaystyle \binom{7}{3} \cdot 3 \cdot 2^3=840$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2009
    Posts
    755
    I didnt quite understand "so now lets see what values of k give 7

    14-3k=7 , k=7/3 and this cant be since k must be +ve integer

    how about 16-3k=7 , k=3 ."

    Does this shows that if the value of r is not a whole number, we have to round the number off and see if the power of x is what we are after?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by Punch View Post
    Find the coefficients of $\displaystyle x^7$ in the expansion of $\displaystyle (1-3x^2)(x^2-\frac{2}{x})^7$

    My attempt:

    to obtain $\displaystyle x^7$, coefficients of $\displaystyle x^5$ and $\displaystyle x^7$ have to be used.

    $\displaystyle \color{blue}x^0$ and $\displaystyle \color{blue}x^7$, $\displaystyle \color{blue}x^2$ and $\displaystyle \color{blue}x^5$

    $\displaystyle T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r$

    the above is fine if you recognise that there is no $\displaystyle \color{blue}x^7$ term when you multiply the binomial expansion by 1

    $\displaystyle \color{blue}x^{14-2r}x^{-r}=x^{14-3r}=x^7$ has no natural solution


    Then the $\displaystyle \color{blue}x^7$ term is obtained by multiplying the binomial expansion by $\displaystyle \color{blue}-3x^2$


    therefore, $\displaystyle 2(7-r)+(-r)=5$
    $\displaystyle r=3$


    OR $\displaystyle 2(7-r)+(-r)=5$
    $\displaystyle r= 2(\frac{1}{3})$ you should have 14-2r-r=14-3r=5, 3r=14-5=9

    Puzzled at arriving at $\displaystyle 2\frac{1}{3}$ for $\displaystyle r$.
    check your 2nd last line simplification again

    .
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2009
    Posts
    755
    alright so can i conclude that I only have to use the term coefficient of x^5 since there is no x^7 term
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by Punch View Post
    alright so can i conclude that I only have to use the term coefficient of x^5 since there is no x^7 term
    Yes, it boils down to finding the $\displaystyle x^5$ term of the binomial.

    Then the coefficient of $\displaystyle x^7$ is $\displaystyle -3(coefficient\ of\ x^5\ from\ the\ binomial\ expansion)$

    This is a shortcut to multiplying out all the terms independently,
    if you only want a single term of the entire expansion of the factors.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Binomial expansion
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Jun 26th 2011, 09:55 AM
  2. [SOLVED] Binomial expansion
    Posted in the Algebra Forum
    Replies: 14
    Last Post: Oct 17th 2010, 09:45 AM
  3. Binomial expansion help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jun 3rd 2009, 06:35 AM
  4. Replies: 6
    Last Post: May 1st 2009, 11:37 AM
  5. Binomial Expansion?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jan 15th 2009, 02:35 PM

Search Tags


/mathhelpforum @mathhelpforum