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    Hi sir, i hope you could answer a few questions of mine:
    For 6ii i drew the correct circle but somehow got the line 'l' wrong and drew the perpendicular bisector of the origin and the centre of the circle
    And then for 6iii i worked out the intersection points by looking at them and checking they were root 5 form the circles centre, but i did this according to the diagram i drew woud i get ecf marks?
    Finally for the last part of the last question, i figured out every root in a unsimplied form eg. 1/(5+4i) but then wrote eg 1/5+1i/4 and split up the fraction for every root which was stupid, how many marks do you think i will get out of 4?
    Id appreciate your help Mr M
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    (Original post by jjjjjjjjjjjj96)
    I'm coming out with the answer (5n + 4) / (6n + 3). Can you explain how you derived the answer for 8iii
    I got the same answer as you

    and I used the summations that Mr M gave in his reply.

    I think we're both wrong? But should get most of the marks.

    The sum to infinity = 1 and then got a common denominator to make it into a single fraction
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    (Original post by duncanjgraham)
    I got the same answer as you

    and I used the summations that Mr M gave in his reply.

    I think we're both wrong? But should get most of the marks.

    The sum to infinity = 1 and then got a common denominator to make it into a single fraction
    \dfrac{n}{3(2n+3)} = \dfrac{n}{6n+9} and \displaystyle\lim_{n\to \infty}{\frac{n}{6n+9}} = \frac{1}{6}
    From this we can do:
    

\displaystyle \sum_n^{\infty}= \sum_1^{\infty} - \sum_{1}^{n-1}

\displaystyle = {\frac{1}{6}} - {\frac{n-1}{6n+3}}

\displaystyle =\frac{(6n+3)-6(n-1)}{36n+18}

\displaystyle =\frac{6n+3-6n+6}{36n+18}

\displaystyle =\frac{9}{36n+18}

\displaystyle =\frac{1}{4n+2}

\displaystyle =\frac{1}{2(2n+1)}
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    (Original post by Nathan20467)
    \dfrac{n}{3(2n+3)} = \dfrac{n}{6n+9} and \displaystyle\lim_{n\to \infty}{\frac{n}{6n+9}} = \frac{1}{6}
    From this we can do:
    

\displaystyle \sum_n^{\infty}= \sum_1^{\infty} - \sum_{1}^{n-1}

\displaystyle = {\frac{1}{6}} - {\frac{n-1}{6n+3}}

\displaystyle =\frac{(6n+3)-6(n-1)}{36n+18}

\displaystyle =\frac{6n+3-6n+6}{36n+18}

\displaystyle =\frac{9}{36n+18}

\displaystyle =\frac{1}{4n+2}

\displaystyle =\frac{1}{2(2n+1)}
    i see ! cheers mate, hadn't learned that before and wasn't able to make that leap on the spot ... rest of the paper was GOOD
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    I made a mistake in 10,mis-factorized 1681 into 11,but the rest are alright,how many marks will I lose for that??
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    (Original post by Nathan20467)
    \dfrac{n}{3(2n+3)} = \dfrac{n}{6n+9} and \displaystyle\lim_{n\to \infty}{\frac{n}{6n+9}} = \frac{1}{6}
    From this we can do:
    

\displaystyle \sum_n^{\infty}= \sum_1^{\infty} - \sum_{1}^{n-1}

\displaystyle = {\frac{1}{6}} - {\frac{n-1}{6n+3}}

\displaystyle =\frac{(6n+3)-6(n-1)}{36n+18}

\displaystyle =\frac{6n+3-6n+6}{36n+18}

\displaystyle =\frac{9}{36n+18}

\displaystyle =\frac{1}{4n+2}

\displaystyle =\frac{1}{2(2n+1)}
    Thanks for putting this on it shows it clearly. I did all of this but so annoyed because didn't halve my first answer which means I got a different answer to this
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    (Original post by Mr M)
    Mr M's OCR (not OCR MEI) Further Pure 1 Answers May 2016


    1. \displaystyle \frac{n(2n+3)(n-1)}{2} (5 marks)


    2. (i) \sqrt{3}-3i (2 marks)

    (ii) \displaystyle \frac{-1+4\sqrt{3} i}{49} (5 marks)


    3. (i) \displaystyle \alpha + \beta = -\frac{1}{k} and \displaystyle \alpha \beta = 1 (1 mark)

    (ii) \displaystyle \frac{1}{k^2} (5 marks)


    4. (i) \begin{pmatrix} 5a-3b & 10 & 0 \end{pmatrix} (2 marks)

    (ii) \begin{pmatrix} 6b-5 \end{pmatrix} (2 marks)

    (iii) \begin{pmatrix} 6a & 12 & 18 \\4a & 8 & 12 \\-a & -2 & -3 \end{pmatrix} (2 marks)


    5. Proof (4 marks)


    6. (i) \displaystyle |z-3-3i|=\sqrt 5 (4 marks)

    (ii) Sketch (2 marks)

    (iii) 2+5i and 4+i (3 marks)


    7. (i) Shear with x axis invariant taking the point (0, 1) to (3, 1) (2 marks)

    (ii) \begin{pmatrix} 0 & -1 \\-1 & 0 \\ \end{pmatrix}

    Reflection in the line y=-x (6 marks)


    8. (i) Show (1 mark)

    (ii) \displaystyle \frac{n}{3(2n+3)} (6 marks)

    (iii) \displaystyle \frac{1}{2(2n+1)} (3 marks)


    9. (i) Show (3 marks)

    (ii) Determinant = 0 when a = 3 so no unique solution. The equations are consistent because 3 x equation 3 - equation 1 = equation 2. (3 marks)


    10. (i) 5+4i and -5-4i (6 marks)

    (ii) Show (1 mark)

    (iii) \displaystyle \pm \frac{5}{41} \pm \frac{4i}{41} (4 marks)
    Sir I made a mistake in 10,mis-factorized 1681 into 11,but the rest are alright,how many marks will I lose for that??
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    (Original post by jonnypiercy)
    what was the final result you had to get for the proof by induction??
    u_1=5 and u_{n+1}=3u_n+2 for n\geq1.

    Prove u_n=2 \times 3^n -1.
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    (Original post by GcseLad-_-)
    Hi sir, i hope you could answer a few questions of mine:
    For 6ii i drew the correct circle but somehow got the line 'l' wrong and drew the perpendicular bisector of the origin and the centre of the circle
    And then for 6iii i worked out the intersection points by looking at them and checking they were root 5 form the circles centre, but i did this according to the diagram i drew woud i get ecf marks?
    Finally for the last part of the last question, i figured out every root in a unsimplied form eg. 1/(5+4i) but then wrote eg 1/5+1i/4 and split up the fraction for every root which was stupid, how many marks do you think i will get out of 4?
    Id appreciate your help Mr M
    6ii) drop 1 and 6iii) drop 1 or 2

    10iii) yuk! drop 1 I expect
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    (Original post by jjjjjjjjjjjj96)
    Please can you show how you got your answer?
    Someone else has done so on page 4.
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    Anybody got an idea what the grade boundaries will be? Do you think mid to high 50's could be a B, because that's what I want!
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    Also, what should the line l, look like on the circle question? Is it perp. to AB through centre of circle?

    Thanks Mr M. for your answers btw, you are a legend!
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    (Original post by Sonnyjimisgod)
    Also, what should the line l, look like on the circle question? Is it perp. to AB through centre of circle?

    Thanks Mr M. for your answers btw, you are a legend!
    I did this using Desmos, I think it's correct, the highlighted points are the points where the circle C, intersects with the line l:

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    (Original post by suriwang)
    Sir I made a mistake in 10,mis-factorized 1681 into 11,but the rest are alright,how many marks will I lose for that??
    Drop 1 or 2 if you are unlucky.
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    (Original post by Sonnyjimisgod)
    Anybody got an idea what the grade boundaries will be? Do you think mid to high 50's could be a B, because that's what I want!
    I don't guess boundaries - it really isn't helpful as people get very disappointed if the result doesn't then turn out well.
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    Am I right in thinking that the quadratic equation was needed for 10iii. I used it and got the answer, but not sure whether it was the correct method
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    Is it ok if you used the quadratic equation for the last question. I got the right answer.
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    With 9 ii.), how many marks would I drop if I stated that the solution was not unique and showed that it was due to the determinant was equal to 3, however stated that the set of equations were inconsistent?
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    (Original post by ChrisWeatherilt)
    Is it ok if you used the quadratic equation for the last question. I got the right answer.
    Is this to find the other root of z^2-18z+1681=0 ? If so, yes.
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    (Original post by deadroad)
    With 9 ii.), how many marks would I drop if I stated that the solution was not unique and showed that it was due to the determinant was equal to 3, however stated that the set of equations were inconsistent?
    Do you mean a = 3 so the determinant = 0? Get 1 mark if you said that.
 
 
 
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