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    H.w : PASCAL Traigle theorm

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    saif alshamery
    عضو مميز
    عضو مميز

    عدد المساهمات : 135
    تاريخ التسجيل : 25/12/2009

    H.w : PASCAL Traigle theorm

    مُساهمة من طرف saif alshamery في الثلاثاء مارس 02, 2010 11:32 pm

    Pascal’s triangle and
    the binomial theorem
    A binomial expression is the sum, or difference, of two terms. For example,
    x + 1, 3x + 2y, a b
    are all binomial expressions. If we want to raise a binomial expression to a power higher than 2
    (for example if we want to find
    (x+1)7) it is very cumbersome to do this by repeatedly multiplying
    x+1 by itself. In this unit you will learn how a triangular pattern of numbers, known as Pascal’s
    triangle
    , can be used to obtain the required result very quickly.
    In order to master the techniques explained here it is vital that you undertake plenty of practice
    exercises so that they become second nature.
    After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
    generate Pascal’s triangle
    expand a binomial expression using Pascal’s triangle
    use the binomial theorem to expand a binomial expression
    Contents
    1. Introduction 2
    2. Pascal’s triangle 2
    3. Using Pascal’s triangle to expand a binomial expression 3
    4. The binomial theorem 6
    1 mc-TY-pascal-2009-1 www.mathcentre.ac.uk c mathcentre June 22, 2009
    1. Introduction
    A binomial expression is the sum, or difference, of two terms. For example,
    x + 1, 3x + 2y, a b
    are all binomial expressions.
    You will be familiar already with the need to expand brackets when squaring such quantities. You
    will know, for example, that
    (x + 1)2 = (x + 1)(x + 1)
    =
    x2 + x + x + 1
    =
    x2 + 2x + 1
    If we want to raise a binomial expression to a power higher than 2 (for example if we want to
    find
    (x + 1)7) it is very cumbersome to do this by repeatedly multiplying x + 1 by itself. In this
    unit you will learn how a triangular pattern of numbers, known as
    Pascal’s triangle, can be used
    to obtain the required result very quickly.
    2. Pascal’s triangle
    We start to generate Pascal’s triangle by writing down the number 1. Then we write a new row
    with the number 1 twice:
    1
    1 1
    We then generate new rows to build a triangle of numbers. Each new row must begin and end
    with a 1:
    1
    1 1
    1 * 1
    1 * * 1
    The remaining numbers in each row are calculated by adding together the two numbers in the
    row above which lie above-left and above-right.
    So, adding the two 1’s in the second row gives 2, and this number goes in the vacant space in
    the third row:
    1
    1 1
    ց ւ
    1 2 1
    1 * * 1
    c mathcentre June 22, 2009 www.mathcentre.ac.uk 2 mc-TY-pascal-2009-1
    The two vacant spaces in the fourth row are each found by adding together the two numbers in
    the third row which lie above-left and above-right:
    1 + 2 = 3, and 2 + 1 = 3. This gives:
    1
    1 1
    ց ւ
    1 2 1
    ց ւ ց ւ
    1 3 3 1
    We can continue to build up the triangle in this way to write down as many rows as we wish.
    The Key Point below shows the first six rows of Pascal’s triangle.
    Key Point
    Pascal’s triangle
    1
    1 1
    1 2 1
    1 3 3 1
    1 4 6 4 1
    1 5 10 10 5 1
    ...
    ...
    Exercise 1
    1. Generate the seventh, eighth, and ninth rows of Pascal’s triangle.
    3. Using Pascal’s triangle to expand a binomial expression
    We will now see how useful the triangle can be when we want to expand a binomial expression.
    Consider the binomial expression
    a + b, and suppose we wish to find (a + b)2.
    We know that
    (a + b)2 = (a + b)(a + b)
    =
    a2 + ab + ba + b2
    = a2 + 2ab + b2
    That is,
    (a + b)2 = 1a2 + 2ab + 1b2
    Observe the following in the final result:
    3
    mc-TY-pascal-2009-1 www.mathcentre.ac.uk c mathcentre June 22, 2009
    1. As we move through each term from left to right, the power of a decreases from 2 down
    to zero.
    2. The power of
    b increases from zero up to 2.
    3. The coefficients of each term, (1, 2, 1), are the numbers which appear in the row of
    Pascal’s triangle beginning 1,2.
    4. The term
    2ab arises from contributions of 1ab and 1ba, i.e. 1ab + 1ba = 2ab. This is the
    link with the way the 2 in Pascal’s triangle is generated; i.e. by adding 1 and 1 in the
    previous row.
    If we want to expand
    (a + b)3 we select the coefficients from the row of the triangle beginning
    1,3: these are 1,3,3,1. We can immediately write down the expansion by remembering that for
    each new term we decrease the power of
    a, this time starting with 3, and increase the power of
    b. So
    (a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3
    which we would normally write as just
    (a + b)3 = a3 + 3a2b + 3ab2 + b3
    Thinking of (a + b)3 as
    (a + b)(a2 + 2ab + b2) = a3 + 2a2b + ab2 + ba2 + 2ab2 + b3 = a3 + 3a2b + 3ab2 + b3
    we note that the term 3ab2, for example, arises from the two terms ab2 and 2ab2; again this is
    the link with the way 3 is generated in Pascal’s triangle - by adding the 1 and 2 in the previous
    row.
    Example
    Suppose we wish to find (a + b)4.
    To find this we use the row beginning 1,4, and can immediately write down the expansion.
    (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
    We can apply the same procedure to expand any binomial expression, even when the quantities
    a and b are more complicated. Consider the following examples.
    Example
    Suppose we want to expand (2x + y)3.
    We pick the coefficients in the expansion from the relevant row of Pascal’s triangle: (1,3,3,1).
    As we move through the terms in the expansion from left to right we remember to decrease the
    power of
    2x and increase the power of y. So,
    (2x + y)3 = 1(2x)3 + 3(2x)2y + 3(2x)1y2 + 1y3
    = 8x3 + 12x2y + 6xy2 + y3
    Example
    Suppose we want to expand (1 + p)4.
    c mathcentre June 22, 2009 www.mathcentre.ac.uk 4 mc-TY-pascal-2009-1
    We pick the coefficients in the expansion from the row of the triangle beginning 1,4; that is
    (1,4,6,4,1). As we move through the terms in the expansion from left to right we remember to
    increase the power of
    p. This example is simpler than the previous one because the first term in
    brackets is 1, and 1 to any power is still 1. So,
    (1 + p)4 = 1(1)4 + 4(1)3p + 6(1)2p2 + 4(1)p3 + 1p4
    = 1 + 4p + 6p2 + 4p3 + p4.
    Either or both of the terms in the binomial expression can be negative. When raising a negative
    number to an even power the result is positive. When raising a negative number to an odd power
    the result is negative. Consider the following example.
    Example
    Expand (3a 2b)5.
    We pick the coefficients in the expansion from the row of Pascal’s triangle beginning 1,5; that is
    1,5,10,10,5,1. Powers of
    3a decrease from 5 as we move left to right. Powers of 2b increase.
    (3a 2b)5 = 1(3a)5 + 5(3a)4(2b) + 10(3a)3(2b)2 + 10(3a)2(2b)3 + 5(3a)(2b)4 + 1(2b)5
    = 243a5 810a4b + 1080a3b2 720a2b3 + 240b4 32b5
    Either or both of the terms could be fractions.
    Example
    Expand 1 +
    2
    x3
    .
    We pick the coefficients in the expansion from the row of Pascal’s triangle (1,3,3,1). Powers of
    2
    x
    increase as we move left to right. Any power of 1 is still 1.
    1 +
    2
    x3
    = 1(1)3 + 3(1)2 2
    x + 3(1)1 2
    x2
    + 12
    x3
    = 1 +
    6
    x
    +
    12
    x2 +
    8
    x3
    Exercises 2
    Use Pascal’s triangle to expand the following binomial expressions:
    1.
    (1 + 3x)2 2. (2 + x)3 3. (1 x)3 4. (1 5x)5
    5. (x + 6)3 6. (a b)7 7. 1 +
    3
    a4
    8. x
    1
    x6
    .
    5
    mc-TY-pascal-2009-1 www.mathcentre.ac.uk c mathcentre June 22, 2009
    4. The binomial theorem
    If we wanted to expand a binomial expression with a large power, e.g. (1 + x)32, use of Pascal’s
    triangle would not be recommended because of the need to generate a large number of rows
    of the triangle. An alternative method is to use the
    binomial theorem. The theorem enables
    us to expand
    (a + b)n in increasing powers of b and decreasing powers of a. We will look at
    expanding expressions of the form
    (a+b)2, (a+b)3, . . . , (a+b)32,. . . , that is when the power is
    a positive whole number. Under certain conditions the theorem can be used when
    n is negative
    or fractional and this is useful in more advanced applications, but these conditions will not be
    studied here.
    Key Point
    The binomial theorem:
    When n is a positive whole number
    (a + b)n = an + nan1b +
    n(n 1)
    2!
    an2b2 +
    n(n 1)(n 2)
    3!
    an3b3
    +
    n(n 1)(n 2)(n 3)
    4!
    an4b4 + . . . + bn
    Note that this is a finite series (that is, it stops after a finite number of terms) and the last term
    is
    bn.
    A simpler form of the theorem is often quoted by taking the special case in which
    a = 1 and
    b = x. It is straightforward to verify that the theorem becomes:
    Key Point
    The binomial theorem:
    When n is a positive whole number
    (1 + x)n = 1 + nx +
    n(n 1)
    2!
    x2 +
    n(n 1)(n 2)
    3!
    x3 +
    n(n 1)(n 2)(n 3)
    4!
    x4 + . . . + xn
    c mathcentre June 22, 2009 www.mathcentre.ac.uk 6 mc-TY-pascal-2009-1
    Example
    We shall apply the binomial theorem to expand (1 + x)2.
    We use the theorem with
    n = 2 and stop when we have written down the term in x2.
    (1 + x)2 = 1 + 2x +
    (2)(2
    1)
    2!
    x2
    = 1 + 2x + x2
    which is the familiar and well-known result.
    Example
    We now apply the binomial theorem to expand (1 + x)3.
    We use the theorem with
    n = 3.
    (1 + x)3 = 1 + 3x +
    (3)(3
    1)
    2!
    x2 +
    (3)(3
    1)(3 2)
    3!
    x3
    = 1 + 3x + 3x2 + x3
    Example
    Suppose we wish to apply the binomial theorem to find the first three terms in ascending powers
    of
    x of (1 + x)32.
    We use the theorem with
    n = 32 and just write down the first three terms.
    (1 + x)32 = 1 + 32x +
    (32)(32
    1)
    2!
    x2
    = 1 + 32x + 496x2 + . . .
    With some ingenuity we can use the theorem to expand other binomial expressions.
    Example
    Suppose we wish to find the first four terms in the expansion of (1 + 1
    3
    y)10.
    We use the theorem, replacing
    x with
    y
    3
    and letting n = 10. This gives
    (1 +
    1
    3
    y)10 = 1 + 10 y
    3 +
    (10)(10
    1)
    2!
    y
    32
    +
    (10)(10
    1)(10 2)
    3!
    y
    33
    + . . .
    = 1 +
    10
    3
    y + 5y2 +
    40
    9
    y3 + . . .
    Example
    Suppose we wish to find the first three terms in the expansion of (3 5z)14.
    We shall apply the binomial theorem in the original form given on page 6 with
    a = 3, b = 5z
    and n = 14.
    (3 5z)14 = 314 + 14(313)(5z) +
    (14)(13)
    2!
    (3
    12)(5z)2
    = 314 (313)70z + (312)2275z2
    = 314 1
    70z
    3
    +
    2275
    9
    z2 . . .
    7 mc-TY-pascal-2009-1 www.mathcentre.ac.uk c mathcentre June 22, 2009
    Exercises 3
    1. Use the binomial theorem to expand (a) (1 + x)4 and (b) (1 + x)5.
    2. Use the binomial theorem to expand
    (1 + 2x)3.
    3. Use the binomial theorem to expand
    (1 3x)4.
    4. Use the binomial theorem to find the first three terms in ascending powers of
    x of (1
    x
    2
    )
    8.
    5. Find the coefficient of
    x5 in the expansion of (1 + 4x)9.
    6. In the expansion of
    (1 x)8 find the coefficient of x7.
    7. Find the first four terms in the expansion of
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    Hero
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    المبدع الذهبي

    عدد المساهمات : 820
    تاريخ التسجيل : 26/02/2010
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    الموقع : ديالى الخالص hero_harb@yahoo.com

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    مُساهمة من طرف Hero في السبت مارس 13, 2010 8:43 am



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