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[FP2 AQA] Help with roots of a cubic and with summations Watch

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    Hello!

    I was working on these questions yesterday:

    https://www.dropbox.com/sh/x2qs0oe7a...IdhTgfx2a?dl=0

    I don't understand why in (3a), the numerator (r+1)(2^{r+1}) - (r+2)(2^r) cancels to r2^r. Can someone explain this to me?

    Also, I don't understand the method the mark scheme uses to prove that (4b) is true, I would have expanded the summation of a^3 + b^3 + y^3 and equated it with 3aby but this would have taken ages for a 3 mark question. Can someone explain the intermediate steps in the mark scheme answer?

    I'm probably being stupid and missing the obvious so help is very much appreciated! Thanks.
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    When you expand the brackets you should and simplify, you should get r2^r+1-r2^r, have you gotten to that stage?


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    (Original post by Haza100)
    Hello!

    I was working on these questions yesterday:

    https://www.dropbox.com/sh/x2qs0oe7a...IdhTgfx2a?dl=0

    I don't understand why in (3a), the numerator (r+1)(2^{r+1}) - (r+2)(2^r) cancels to r2^r. Can someone explain this to me?
    \displaystyle (r+1)2^{r+1} = r2^{r+1} + 2^{r+1}

    \displaystyle (r+2)(2^r) = r2^r + 2^1 \times 2^r = r2^r + 2^{r+1}

    \displaystyle (r+1)2^r - (r+2)2^r = r2^{r+1} + 2^{r+1} - r2^r - 2^{r+1}= 2r2^r +2^{r+1} - r2^r - 2^{r+1}

    Can you see how to finish this off?
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    (Original post by drandy76)
    When you expand the brackets you should and simplify, you should get r2^r+1-r2^r, have you gotten to that stage?


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    Only just realised that  2(2^r) --> 2^{r+1}. Duh! Guess I worked for too long yesterday haha.
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    (Original post by Zacken)
    \displaystyle (r+1)2^{r+1} = r2^{r+1} + 2^{r+1}

    \displaystyle (r+2)(2^r) = r2^r + 2^1 \times 2^r = r2^r + 2^{r+1}

    \displaystyle (r+1)2^r - (r+2)2^r = r2^{r+1} + 2^{r+1} - r2^r - 2^{r+1}= 2r2^r +2^{r+1} - r2^r - 2^{r+1}

    Can you see how to finish this off?
    Yeah thanks, I stupidly missed 2^1 == 2. Thanks
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    (Original post by Haza100)

    Also, I don't understand the method the mark scheme uses to prove that (4b) is true, I would have expanded the summation of a^3 + b^3 + y^3 and equated it with 3aby but this would have taken ages for a 3 mark question. Can someone explain the intermediate steps in the mark scheme answer?
    So you know that a, b, c are roots of the equation (I'm using those in place of \alpha, \beta and \gamma respectively)

    So you know that:

    a^3 + pa + q = 0

    b^3 + pb + q = 0

    c^3 + pc + q = 0

    You can now add all three equations together to get:

    a^3 + b^3 + c^3 + p(a+b+c) + 3q = 0

    You've found a+b+c already in the above parts, so plug that in. Can you now see how to finish off?
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    (Original post by Haza100)
    Only just realised that  2(2^r) --> 2^{r+1}. Duh! Guess I worked for too long yesterday haha.
    For 4(b) it looks like they've done an equivalent to what you did, but used summation formulas for conciseness


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    (Original post by Zacken)
    So you know that a, b, c are roots of the equation (I'm using those in place of \alpha, \beta and \gamma respectively)

    So you know that:

    a^3 + pa + q = 0

    b^3 + pb + q = 0

    c^3 + pc + q = 0

    You can now add all three equations together to get:

    a^3 + b^3 + c^3 + p(a+b+c) + 3q = 0

    You've found a+b+c already in the above parts, so plug that in. Can you now see how to finish off?
    Ah yes, thank you so much.
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    (Original post by Haza100)
    Ah yes, thank you so much.
    Glad I helped.
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    am I too late?
 
 
 
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