Mr M's OCR (not OCR MEI) Further Pure 1 Answers May 2016

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

$\dfrac{n}{3(2n+3)} = \dfrac{n}{6n+9}$ and $\displaystyle\lim_{n\to \infty}{\frac{n}{6n+9}} = \frac{1}{6}$
From this we can do:
$[br]\displaystyle \sum_n^{\infty}= \sum_1^{\infty} - \sum_{1}^{n-1}[br]\displaystyle = {\frac{1}{6}} - {\frac{n-1}{6n+3}}[br]\displaystyle =\frac{(6n+3)-6(n-1)}{36n+18}[br]\displaystyle =\frac{6n+3-6n+6}{36n+18}[br]\displaystyle =\frac{9}{36n+18}[br]\displaystyle =\frac{1}{4n+2}[br]\displaystyle =\frac{1}{2(2n+1)}[br]$

$\dfrac{n}{3(2n+3)} = \dfrac{n}{6n+9}$ and $\displaystyle\lim_{n\to \infty}{\frac{n}{6n+9}} = \frac{1}{6}$
From this we can do:
$[br]\displaystyle \sum_n^{\infty}= \sum_1^{\infty} - \sum_{1}^{n-1}[br]\displaystyle = {\frac{1}{6}} - {\frac{n-1}{6n+3}}[br]\displaystyle =\frac{(6n+3)-6(n-1)}{36n+18}[br]\displaystyle =\frac{6n+3-6n+6}{36n+18}[br]\displaystyle =\frac{9}{36n+18}[br]\displaystyle =\frac{1}{4n+2}[br]\displaystyle =\frac{1}{2(2n+1)}[br]$

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)

what was the final result you had to get for the proof by induction??

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

Please can you show how you got your answer?

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!

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??

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!

Is it ok if you used the quadratic equation for the last question. I got the right answer.

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.