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    I'm trying to calculate and I'm using and expanding it to do so. I have got . I'm using x = 0.2. The question requires working it out to 2dp without a calculator. The calculations seem extremely laborious, so does anyone have any tips that will make it easier or a shortcut?

    Thanks
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    (Original post by Namige)
    I'm trying to calculate and I'm using
    and expanding it to do so. I have got The question requires working it out to 2dp without a calculator. The calculations seem extremely laborious, so does anyone have any tips that will make it easier or a shortcut?

    Thanks
    Your expansion of (1+x)^4 is incorrect

    Use Pascal's triangle - 1, 4, 6, 4,1
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    (Original post by Indeterminate)
    Your expansion of (1+x)^4 is incorrect

    Use Pascal's triangle - 1, 4, 6, 4,1
    Sorry, my mistake and why do I need pascal's triangle for this?
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    (Original post by Namige)
    Sorry, my mistake and why do I need pascal's triangle for this?
    Pascal's triangle gives the coefficients of each power of x and y in an expansion of a bracket, (x+y), to an integer (whole number) power.

    You look at the line that is the power you're dealing with plus one So, for example, the 3rd line gives the coefficients for the power 2 - 1,2,1

    (x+y)^2 = x^2 + 2xy + y^2

    The fourth line gives them for the power 3 - 1,3,3,1

    (x+y)^3 = x^3 + 3x^2 y + 3xy^2 +y^3

    etc.

    Notice the symmetry of the powers of x and y in the above expansions.
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    (Original post by Indeterminate)
    Pascal's triangle gives the coefficients of each power of x and y in an expansion of a bracket, (x+y), to an integer (whole number) power.

    You look at the line that is the power you're dealing with plus one So, for example, the 3rd line gives the coefficients for the power 2 - 1,2,1

    (x+y)^2 = x^2 + 2xy + y^2

    The fourth line gives them for the power 3 - 1,3,3,1

    (x+y)^3 = x^3 + 3x^2 y + 3xy^2 +y^3

    etc.

    Notice the symmetry of the powers of x and y in the above expansions.
    I've already expanded it?
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    (Original post by Namige)
    I've already expanded it?
    Using the 5th line - 1,4,6,4,1

    (x+y)^4 = x^4 + 4x^3 y + 6x^2 y^2 + 4xy^3 + y^4

    Again, notice the symmetry of the powers.

    How does this simplify in your case?
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    (Original post by Indeterminate)
    Using the 5th line - 1,4,6,4,1

    (x+y)^4 = x^4 + 4x^3 y + 6x^2 y^2 + 4xy^3 + y^4

    Again, notice the symmetry of the powers.

    How does this simplify in your case?
    Thanks
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    (Original post by Indeterminate)
    Using the 5th line - 1,4,6,4,1

    (x+y)^4 = x^4 + 4x^3 y + 6x^2 y^2 + 4xy^3 + y^4

    Again, notice the symmetry of the powers.

    How does this simplify in your case?
    How come I'm told by my teachers to turn it into (1+x)^n before expanding?
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    (Original post by Namige)
    Thanks
    Ah yes, so I'm assuming that 1 (instead of 4) was a typo in the first post, because your expansion for (4+x)^4 is correct.

    For (4.2)^4

    there isn't a shortcut I'm afraid unless you use a calculator (which you're not allowed to do).

    Mental arithmetic!!
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    (Original post by Namige)
    How come I'm told by my teachers to turn it into (1+x)^n before expanding?
    For integer (whole number powers), you can do it either way.

    Maybe they think you're less likely to make an error
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    (Original post by Indeterminate)
    For integer (whole number powers), you can do it either way.

    Maybe they think you're less likely to make an error
    Can you tell me why pascal's triangle does not work for natural number powers, but the general formula ( n(n-1)/2! ...) works?
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    (Original post by Namige)
    Can you tell me why pascal's triangle does not work for natural number powers, but the general formula ( n(n-1)/2! ...) works?
    Natural numbers are whole numbers, and it clearly works for them. I think you mean fractions.
    If you do, notice that there are infinitely many terms in an expansion to a fractional power, so we can't possibly state every coefficient.


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