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    a) Expand (1 3x)^8 in ascending powers of x, up to and including the term in x^3.b) Showing your working clearly, use your expansion to find, to 5 significant figures, an approximation for 1.03^8. please can someone help me answer this??
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    (Original post by bobthebob123)
    a) Expand (1 3x)^8 in ascending powers of x, up to and including the term in x^3.b) Showing your working clearly, use your expansion to find, to 5 significant figures, an approximation for 1.03^8. please can someone help me answer this??
    is it (1*3x)^8 or (1+3x)^8 or (1-3x)^8?
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    (Original post by bobthebob123)
    a) Expand (1 3x)^8 in ascending powers of x, up to and including the term in x^3.b) Showing your working clearly, use your expansion to find, to 5 significant figures, an approximation for 1.03^8. please can someone help me answer this??
    part b, once you have if expansion up to x^3 you will equate what's in the brackets to 1.03 and solve for x Then replace this value of x in Ur expansion and calculate it. you will get an approximate
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    (Original post by brainmaster)
    is it (1*3x)^8 or (1+3x)^8 or (1-3x)^8?
    oh sorry it is (1+3x)^8
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    (Original post by bobthebob123)
    oh sorry it is (1 3x)^8
    use the formula;
    1 nx ((n (n-1)/2!)(x)^2 ((n (n-1)(n-2)/3!)(x)^3.....
    And part b)
    equate the 1 3x to 1.03 and find value of x and use this value in your expansion to get the approximate.
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    (Original post by brainmaster)
    the expansion would be;
    1+8 (3x)
    let me know if this was helpful😊
    Please edit your post - we don't give answers in the Maths forum we just give hints.
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    (Original post by Muttley79)
    Please edit your post - we don't give answers in the Maths forum we just give hints.
    I'm New here but thank I'll remember this....is the post okay now?
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    (Original post by brainmaster)
    I'm New here but thank I'll remember this....is the post okay now?
    Thank you, yes

    Do stay around and help - we find hints help more than doing a question for someone.
 
 
 
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