Edexcel A-Level Mathematics A2 Fp2 2 Series Questions

$$\frac18\sum_{r=1}^{n}\left(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\right)$$$$=\frac18\left(1-\frac13-\frac{1}{2n+1}+\frac{1}{2n+3}\right)$$$$=\frac18\left(\frac{2(2n+1)(2n+3)-3(2n+3)+3(2n+1)}{3(2n+1)(2n+3)}\right)$$$$=\frac{n(n+2)}{3(2n+1)(2n+3)}$$

$$\begin{aligned} r=n-1 \quad \frac{1}{n-2}-\frac{2}{n-1}+\frac{1}{n} r=n \quad \frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1} \end{aligned}$$$\sum_{r=2}^{n}\left(\frac{1}{r-1}-\frac{2}{r}+\frac{1}{r+1}\right)=\left(1-\frac{2}{2}+\frac{1}{2}+\frac{1}{n}-\frac{2}{n}+\frac{1}{n+1}\right)$$\frac{1}{2} \sum_{r=1}^{n} \frac{2}{r(r+1)(r-1)}=\frac{1}{2} \times\left(\frac{1}{2}-\frac{1}{n}+\frac{1}{n+1}\right)=\frac{n^{2}+n-2}{4 n(n+1)}$

(a)

Attempt PFs by any valid method (by implication if 3 correct fractions seen) A1 any 2 fractions correct;
A1 third fraction correct

Method of differences with at least 3 terms at start and 2 at end OR 2 at start and 3 at end. Must start at 2 and end at n One M mark for the initial terms and a second for the final. Last lines may be missing k /(n-1) and c /(n-2) These 2 M marks may be implied by a correct extraction of terms. If starting from 1, M0M1 can be awarded.
Extract the remaining terms. 1-2 / 2 may be missing and 1 / n-2 / n may be combined Include the 1 / 2 and attempt a common denominator of the required form. Depends on both previous M marks
$$\frac{n^{2}+n-2}{4 n(n+1)}$$}

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Number

Alternative for (b) using integration by parts (Very rare but may be seen)
$$\begin{aligned} & \text { Area }=\frac{1}{2} \int_{0}^{\frac{\pi}{3}}(1+\cos \theta)^{2} \mathrm{~d} \theta & =\frac{1}{2}\left[\int(1+2 \cos \theta) \mathrm{d} \theta+\int \cos ^{2} \theta \mathrm{~d} \theta\right] \end{aligned}$$$=\frac{1}{2}\left[\int(1+2 \cos \theta) \mathrm{d} \theta+\cos \theta \sin \theta+\int \sin ^{2} \theta \mathrm{~d} \theta\right]$$$=\frac{1}{2}\left[\theta+2 \sin \theta+\sin \theta \cos \theta+\int\left(1-\cos ^{2} \theta\right) \mathrm{d} \theta\right]_{0}^{\frac{\pi}{3}}$$$=\frac{1}{2}\left[\theta+2 \sin \theta+\frac{1}{2}(\sin \theta \cos \theta+\theta)\right]_{0}^{\frac{\pi}{3}}$$=\frac{\pi}{4}+\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{16}=\frac{\pi}{4}+\frac{9 \sqrt{3}}{16}$
М1
Use of Area $=\frac{1}{2} \int r^{2} \mathrm{~d} \theta$ with $r=1+\cos \theta$, limits not needed.

Attempt $(1+\cos \theta)^{2}$ (minimum accepted is $\left(1+k \cos \theta+\cos ^{2} \theta\right)$ ) and attempt first stage of $\int \cos ^{2} \theta \mathrm{~d} \theta$ by parts. Reach $\int \cos ^{2} \theta \mathrm{~d} \theta=\cos \theta \sin \theta \pm \int \sin ^{2} \theta \mathrm{~d} \theta$ Limits not needed Correct so far. Limits not needed.

Attempt to integrate all terms. $\int(1+2 \cos \theta) \mathrm{d} \theta$ and attempt to complete $\int \cos ^{2} \theta \mathrm{~d} \theta$ using Pythagoras identity. Limits not needed. Depends on the previous M mark As main scheme

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Scheme

$$\begin{aligned} \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{4}{y}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{2}-3 \frac{\mathrm{~d}^{3} y}{\mathrm{~d} x^{3}}=-\frac{4}{y^{2}}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{3}+\frac{8}{y} \times \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} \times \frac{\mathrm{d} y}{\mathrm{~d} x} \frac{\mathrm{~d}^{3} y}{\mathrm{~d} x^{3}}=-\frac{4}{y^{2}}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{3}+\frac{8}{y}\left(\frac{4}{y}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{2}-3\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right) \frac{\mathrm{d}^{3} y}{\mathrm{~d} x^{3}}=\frac{28}{y^{2}}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{3}-\frac{24}{y}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right) * \end{aligned}$$

studentarts
(5)

ALT

$$\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x} \frac{\mathrm{~d}^{2} y}{\mathrm{~d} x^{2}}+y \frac{\mathrm{~d}^{3} y}{\mathrm{~d} x^{3}}-8 \frac{\mathrm{~d} y}{\mathrm{~d} x} \times \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+3 \frac{\mathrm{~d} y}{\mathrm{~d} x}=0 \frac{\mathrm{~d}^{3} y}{\mathrm{~d} x^{3}}=\frac{1}{y}\left(7 \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)\left(\frac{4}{y}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{2}-3\right)-\frac{3}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x} \frac{\mathrm{~d}^{3} y}{\mathrm{~d} x^{3}}=\frac{28}{y^{2}}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{3}-\frac{24}{y}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right) * \end{aligned}$$$$\mathrm{A1}^{*}$$