Results 1 to 11 of 11

Math Help - any one help.....so hard to solve...!!!

  1. #1
    Newbie sonymd23's Avatar
    Joined
    Jul 2006
    Posts
    22

    any one help.....so hard to solve...!!!

    1. predict the result of Alternately adding and subtracting the squares of the terms in the nth row of Pascal's Triangle.

    2. given the last four terms of a row of Pascal's Triangle are : 816 , 153 , 18 , 1. determine the last four terms of the next row

    3. the number 23040 can be factored into prime divisors as 2x2x2x3x3x5x8x8. determine the number of even divisors of 23040


    ^ thats the question....can any one solve it....thnx......show ur steps and remember explain...... thnx very much
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by sonymd23
    1. predict the result of Alternately adding and subtracting the squares of the terms in the nth row of Pascal's Triangle.
    Powers of 2, i.e. 2^n
    Since the sum of the rows of a Pascal triangle are,
    \sum_{k=0}^n{n \choose 0}+{n \choose 1}+...+{n \choose n}=(1+1)^n=2^n
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by sonymd23
    2. given the last four terms of a row of Pascal's Triangle are : 816 , 153 , 18 , 1. determine the last four terms of the next row
    The 2nd to last term is,
    {n \choose 1}=n=18
    Therefore the k-th to last term is,
    {18 \choose k-1}
    Now evaluate it this for k=5,6,7,8
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by sonymd23
    3. the number 23040 can be factored into prime divisors as 2x2x2x3x3x5x8x8. determine the number of even divisors of 23040
    Its prime factorization is,
    2^9\cdot 3^2\cdot 5
    Note any divisor must be a form of,
    N=2^a\cdot 3^b\cdot 5^c
    Where,
    0\leq a\leq 9,0\leq b\leq 2, 0\leq c\leq 1But since we want N is even we need that a\geq 1
    Thus, in total we have 9 possibilities for a 3 possibilities for c and 2 possibilities for c
    Thus, by the fundamental counting princple in total,
    9\times 3\times 2=54
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by ThePerfectHacker
    Powers of 2, i.e. 2^n
    Since the sum of the rows of a Pascal triangle are,
    \sum_{k=0}^n{n \choose 0}+{n \choose 1}+...+{n \choose n}=(1+1)^n=2^n
    i think that squares of terms are being added or subtracted, not the terms.

    Malay
    Last edited by malaygoel; August 7th 2006 at 03:08 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Powers of 2, i.e. 2^n
    Since the sum of the rows of a Pascal triangle are,
    \sum_{k=0}^n{n \choose 0}+{n \choose 1}+...+{n \choose n}=(1+1)^n=2^n
    Except (as far as I can tell) the question asks for:

    <br />
\sum_{k=0}^n (-1)^k{n \choose k}^2

    RonL
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by ThePerfectHacker
    The 2nd to last term is,
    {n \choose 1}=n=18
    Therefore the k-th to last term is,
    {18 \choose k-1}
    Now evaluate it this for k=5,6,7,8
    terms of next row are wanted, not the next terms of the same row

    Malay
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    The 2nd to last term is,

    {n \choose 1}=n=18

    Therefore the k-th to last term is,

    {18 \choose k-1}

    Now evaluate it this for k=5,6,7,8
    The 2nd to last terms is:

    {n \choose 1}=n=18,

    so this is the 19th row, and the next row is the 20th, and so the last four terms are:

    <br />
{19 \choose 19-k},\ \ \ k=3,\ 2,\ 1,\ 0<br />

    RonL
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Forgive me, I have a disease which makes me see problems diffrently.

    In that case the answer is zero, cuz
    Since,
    {n \choose k}={n \choose n-k}

    (I did not fully work it out so It might be based on even or odd.)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Forgive me, I have a disease which makes me see problems diffrently.

    In that case the answer is zero, cuz
    Since,
    {n \choose k}={n \choose n-k}

    (I did not fully work it out so It might be based on even or odd.)
    Great mystery of the day - which question are you refering to?
    (Rhetorical question, I know but a casual reader may get lost)

    Its zero only when k is even surly?

    Code:
                            1
                     +1    -4     +1             !=0
                  +1    -9    +9     -1           =0
               +1   -16   +36    -16    +1       !=0
    and so on
    RonL
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,713
    Thanks
    633
    Hello, sonymd23!

    #3 has an incorrect phrase in it . . . Is there a typo?
    . . And it is easier if you know a particular theorem.


    3. The number 23040 can be factored into prime divisors as:
    . . . 2\times2\times2\times3\times3\times5\times8\times8
    . . . but these are not prime divisors
    Determine the number of even divisors of 23040.

    Theorem

    If the prime factorization of N is: N \:=\:2^a3^b5^c\hdots
    . . then the number of divisors of N is: (a + 1)(b + 1)(c + 1) \hdots
    . . (Add one to each exponent, and multiply.)
    This includes the divisors 1 and Nitself.


    We have: . 23040\:=\:2^9\cdot3^2\cdot5^1

    Hence, 23040 has: 10\cdot3\cdot2 = 60 divisors.


    How many of these are odd divisors?
    . . An odd divisor must be a product of odd factors.
    Odd divisors would come from: 3^2\cdot5
    . . which has (2+1)(1+1) = 6 divisors.


    Therefore, 23040 has: 60 - 6 \,=\,\boxed{54} even divisors
    . . which verifies TPHacker's solution.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Coin tossing and Cards Experiment Which I Find Hard To Solve
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: April 13th 2010, 04:34 PM
  2. Some hard limits I can't solve
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 6th 2009, 04:29 AM
  3. A integral problem hard to solve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 30th 2009, 10:32 AM
  4. Replies: 3
    Last Post: May 31st 2008, 05:45 AM
  5. Replies: 0
    Last Post: February 11th 2008, 03:22 AM

Search Tags


/mathhelpforum @mathhelpforum