Problems
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1991 Paper 3 Q1
- Evaluate \[ \sum_{r=1}^{n}\frac{6}{r(r+1)(r+3)}. \]
- Expand \(\ln(1+x+x^{2}+x^{3})\) as a series in powers of \(x\), where \(\left|x\right|<1\), giving the first five non-zero terms and the general term.
- Expand \(\mathrm{e}^{x\ln(1+x)}\) as a series in powers of \(x\), where \(-1 < x\leqslant1\), as far as the term in \(x^{4}\).
Solution:
- \begin{align*} && \frac{6}{r(r+1)(r+3)} &= \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \\ \Rightarrow && \sum_{r=1}^n \frac{6}{r(r+1)(r+3)} &= \sum_{r=1}^n \l \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \r \\ &&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=1}^n \frac{3}{r+1} + \sum_{r=1}^n \frac{1}{r+3} \\ &&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=2}^{n+1} \frac{3}{r} + \sum_{r=3}^{n+2} \frac{1}{r} \\ &&& = \frac{2}{1} + \frac{2}{2} - \frac{3}{2} - \frac{3}{n+1} + \frac{1}{n+1} + \frac{1}{n+2} \\ &&& = \frac{3}{2} - \frac{2}{n+1} + \frac{1}{n+2} \end{align*}
- \begin{align*} && \ln (1 + x+ x^2 + x^3) &= \ln \l \frac{1-x^4}{1-x} \r \\ &&&= \ln (1-x^4) - \ln(1-x) \\ &&&= \sum_{k=1}^{\infty} -\frac{x^{4k}}{k} - \sum_{k=1}^{\infty} - \frac{x^k}{k} \\ &&&= x + \frac12x^2+\frac13x^3-\frac34x^4+\frac15x^5 + \cdots \\ &&&= \sum_{k=1}^{\infty}a_k x^k \end{align*} Where \(a_k = \frac{1}{k}\) if \(k \neq 0 \pmod{4}\) otherwise \(a_k = -\frac{3}{k}\) if \(k \equiv 0 \pmod{4}\)
- \begin{align*} \exp(x \ln (1+x) ) &= \exp\l x \l x-\frac12x^2+\frac13x^3-\cdots \r \r \\ &= \exp\l x^2-\frac12x^3+\frac13x^4 \r \\ &= 1 + \l x^2-\frac12x^3+\frac13x^4 \r + \frac12 \l x^2-\frac12x^3+\frac13x^4 \r^2 + \cdots \\ &= 1 + x^2-\frac12x^3+\frac13x^4 + \frac12x^4 + \cdots \\ &= 1 + x^2 -\frac12x^3+\frac56x^4+\cdots \end{align*}
1991 Paper 3 Q2
The distinct points \(P_{1},P_{2},P_{3},Q_{1},Q_{2}\) and \(Q_{3}\) in the Argand diagram are represented by the complex numbers \(z_{1},z_{2},z_{3},w_{1},w_{2}\) and \(w_{3}\) respectively. Show that the triangles \(P_{1}P_{2}P_{3}\) and \(Q_{1}Q_{2}Q_{3}\) are similar, with \(P_{i}\) corresponding to \(Q_{i}\) (\(i=1,2,3\)) and the rotation from \(1\) to \(2\) to \(3\) being in the same sense for both triangles, if and only if \[ \frac{z_{1}-z_{2}}{z_{2}-z_{3}}=\frac{w_{1}-w_{2}}{w_{1}-w_{3}}. \] Verify that this condition may be written \[ \det\begin{pmatrix}z_{1} & z_{2} & z_{3}\\ w_{1} & w_{2} & w_{3}\\ 1 & 1 & 1 \end{pmatrix}=0. \]
- Show that if \(w_{i}=z_{i}^{2}\) (\(i=1,2,3\)) then triangle \(P_{1}P_{2}P_{3}\) is not similar to triangle \(Q_{1}Q_{2}Q_{3}.\)
- Show that if \(w_{i}=z_{i}^{3}\) (\(i=1,2,3\)) then triangle \(P_{1}P_{2}P_{3}\) is similar to triangle \(Q_{1}Q_{2}Q_{3}\) if and only if the centroid of triangle \(P_{1}P_{2}P_{3}\) is the origin. {[}The centroid of triangle \(P_{1}P_{2}P_{3}\) is represented by the complex number \(\frac{1}{3}(z_{1}+z_{2}+z_{3})\).{]}
- Show that the triangle \(P_{1}P_{2}P_{3}\) is equilateral if and only if \[ z_{2}z_{3}+z_{3}z_{1}+z_{1}z_{2}=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}. \]
1991 Paper 3 Q3
The function \(\mathrm{f}\) is defined for \(x<2\) by \[ \mathrm{f}(x)=2| x^{2}-x|+|x^{2}-1|-2|x^{2}+x|. \] Find the maximum and minimum points and the points of inflection of the graph of \(\mathrm{f}\) and sketch this graph. Is \(\mathrm{f}\) continuous everywhere? Is \(\mathrm{f}\) differentiable everywhere? Find the inverse of the function \(\mathrm{f}\), i.e. expressions for \(\mathrm{f}^{-1}(x),\) defined in the various appropriate intervals.
Solution: \[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \] Therefore the absolute value terms will change behaviour at \(x = -1, 0, 1\). Then \begin{align*} f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) & x \leq -1 \\ 2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\ -2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\ 2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\ &= \begin{cases} x^2-4x-1 & x \leq -1 \\ 3x^2+1& -1 < x \leq 0 \\ -5x^2+1& 0 < x \leq 1 \\ x^2-4x-1 & 1 < x\end{cases} \\ \\ f'(x) &= \begin{cases} 2x-4 & x <-1 \\ 6x & -1 < x < 0 \\ -10x & 0 < x < 1 \\ 2x-4 & 1 < x\end{cases} \\ \end{align*} Therefore \(f'(x) = 0 \Rightarrow x = 0, 2\) and so we should check all the turning points. Therefore the minimum is \(x = 2, y = -5\), maximum is \(x = -2, y = 11\) (assuming the range is actually \(|x| < 2\). There is a point of inflection at \(x = 0, y = 1\).
1991 Paper 3 Q4
The point \(P\) moves on a straight line in three-dimensional space. The position of \(P\) is observed from the points \(O_{1}(0,0,0)\) and \(O_{2}(8a,0,0).\) At times \(t=t_{1}\) and \(t=t_{1}'\), the lines of sight from \(O_{1}\) are along the lines \[ \frac{x}{2}=\frac{z}{3},y=0\quad\mbox{ and }\quad x=0,\frac{y}{3}=\frac{z}{4} \] respectively. At times \(t=t_{2}\) and \(t=t_{2}'\), the lines of sight from \(O_{2}\) are \[ \frac{x-8a}{-3}=\frac{y}{1}=\frac{z}{3}\quad\mbox{ and }\quad\frac{x-8a}{-4}=\frac{y}{2}=\frac{z}{5} \] respectively. Find an equation or equations for the path of \(P\).
1991 Paper 3 Q5
The curve \(C\) has the differential equation in polar coordinates \[ \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r=5\sin3\theta,\qquad\text{for }\quad\frac{\pi}{5}\leqslant\theta\leqslant\frac{3\pi}{5}, \] and, when \(\theta=\dfrac{\pi}{2},\) \(r=1\) and \(\dfrac{\mathrm{d}r}{\mathrm{d}\theta}=-2.\) Show that \(C\) forms a closed loop and that the area of the region enclosed by \(C\) is \[ \frac{\pi}{5}+\frac{25}{48}\left[\sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right)\right]. \]
Solution: First we seek the complementary function. \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 0 \\ \Rightarrow && r &= A \sin 2\theta + B \cos 2 \theta \end{align*} Next we seek a particular integral, of the form \(r = C \sin 3 \theta\). \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 5 \sin 3 \theta \\ \Rightarrow && -9C \sin 3 \theta + 4C \sin 3 \theta &= 5 \sin 3 \theta \\ \Rightarrow && C &= -1 \\ \end{align*} So our general solution is \(A \sin 2\theta + B \cos 2 \theta -\sin 3 \theta\). Plugging in boundary conditions we obtain: \begin{align*} \theta = \frac{\pi}{2}, r = 1: &&1 &= -B +1 \\ \Rightarrow && B &= 0 \\ \theta = \frac{\pi}{2}, \frac{\d r}{\d \theta} = -2: && -2 &= -2A \\ \Rightarrow && A &= 1 \end{align*} So the general solution is \(r = \sin 2 \theta - \sin 3 \theta = 2 \sin \left ( \frac{-\theta}{2} \right) \cos \left (\frac{5 \theta}{2} \right)\) First notice that for \(\theta \in \left [\frac{\pi}{5}, \frac{3 \pi}{5} \right]\) this is positive, and it is zero on the end points, therefore we are tracing out a a loop. The area of the loop will be: \begin{align*} A &= \int_{\pi/5}^{3\pi/5} \frac12 \left ( \sin 2 \theta - \sin 3 \theta \right)^2 \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \sin^2 2\theta + \sin^2 3 \theta - 2 \sin 2 \theta \cos 3 \theta \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \frac{1-2 \cos 4 \theta}{2} + \frac{1-2 \cos6 \theta}{2} - \sin5 \theta-\cos\theta \d \theta \\ &= \frac12 \left [\theta - \frac14 \sin 4 \theta-\frac16 \sin 6 \theta + \frac15 \cos 5 \theta - \sin \theta \right]_{\pi/5}^{3\pi/5} \\ &= \frac{\pi}{5} +\frac{25}{48}\left [ \sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right) \right] \end{align*}
1991 Paper 3 Q6
The transformation \(T\) from \(\binom{x}{y}\) to \(\binom{x'}{y'}\) in two-dimensional space is given by \[ \begin{pmatrix}x'\\ y' \end{pmatrix}=\begin{pmatrix}\cosh u & \sinh u\\ \sinh u & \cosh u \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix}, \] where \(u\) is a positive real constant. Show that the curve with equation \(x^{2}-y^{2}=1\) is transformed into itself. Find the equations of two straight lines through the origin which transform into themselves. A line, not necessary through the origin, which has gradient \(\tanh v\) transforms under \(T\) into a line with gradient \(\tanh v'\). Show that \(v'=v+u\). The lines \(\ell_{1}\) and \(\ell_{2}\) with gradients \(\tanh v_{1}\) and \(\tanh v_{2}\) transform under \(T\) into lines with gradients \(\tanh v_{1}'\) and \(\tanh v_{2}'\) respectively. Find the relation satisfied by \(v_{1}\) and \(v_{2}\) that is the necessary and sufficient for \(\ell_{1}\) and \(\ell_{2}\) to intersect at the same angle as their transforms. In the case when \(\ell_{1}\) and \(\ell_{2}\) meet at the origin, illustrate in a diagram the relation between \(\ell_{1}\), \(\ell_{2}\) and their transforms.
1991 Paper 3 Q7
- Prove that \[ \int_{0}^{\frac{1}{2}\pi}\ln(\sin x)\,\mathrm{d}x=\int_{0}^{\frac{1}{2}\pi}\ln(\cos x)\,\mathrm{d}x=\tfrac{1}{2}\int_{0}^{\frac{1}{2}\pi}\ln(\sin2x)\,\mathrm{d}x-\tfrac{1}{4}\pi\ln2 \] and \[ \int_{0}^{\frac{1}{2}\pi}\ln(\sin2x)\,\mathrm{d}x=\tfrac{1}{2}\int_{0}^{\pi}\ln(\sin x)\,\mathrm{d}x=\int_{0}^{\frac{1}{2}\pi}\ln(\sin x)\,\mathrm{d}x. \] Hence, or otherwise, evaluate \({\displaystyle \int_{0}^{\frac{1}{2}\pi}\ln(\sin x)\,\mathrm{d}x.}\) You may assume that all the integrals converge.
- Given that \(\ln u< u\) for \(u\geqslant1\) deduce that \[ \tfrac{1}{2}\ln x < \sqrt{x}\qquad\mbox{ for }\quad x\geqslant1. \] Deduce that \(\dfrac{\ln x}{x}\rightarrow0\) as \(x\rightarrow\infty\) and that \(x\ln x\rightarrow0\) as \(x\rightarrow0\) through positive values.
- Using the results of parts (i) and (ii), or otherwise, evaluate \({\displaystyle \int_{0}^{\frac{1}{2}\pi}x\cot x\,\mathrm{d}x.}\)
Solution:
- \begin{align*} u = \frac{\pi}{2} - x :&& \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \int_{\frac12\pi}^0 \ln (\cos u) (- 1)\d u \\ &&&= \int_0^{\frac12 \pi} \ln (\cos x) \d x \\ \Rightarrow && 2 \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \int_0^{\tfrac12 \pi} \ln (\sin x) \d x +\int_0^{\tfrac12 \pi} \ln (\cos x) \d x \\ &&&= \int_0^{\tfrac12 \pi}\left (\ln (\sin x)+ \ln (\cos x) \right) \d x \\ &&&= \int_0^{\frac12 \pi} \ln \left (\frac12 \sin 2x \right) \d x \\ &&&= \int_0^{\frac12 \pi} \left ( \ln \left (\sin 2x \right) - \ln 2 \right)\d x \\ &&&= \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x - \frac{\pi}{2} \ln 2\\ \Rightarrow && \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \frac12 \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x - \frac{\pi}{4} \ln 2 \end{align*} \begin{align*} u = 2x, \d u = 2 \d x && \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x &= \int_0^{\pi} \ln (\sin u) \frac12 \d u \\ &&&= \frac12 \int_0^{\pi} \ln (\sin u) \d u \\ &&&=\frac12 \left ( \int_0^{\pi/2} \ln (\sin u) \d u + \int_{\pi/2}^{\pi} \ln (\sin u) \d u \right)\\ &&&= \int_0^{\pi/2} \ln (\sin u) \d u \\ \Rightarrow && I &= \frac12 I - \frac14 \pi \ln 2 \\ \Rightarrow && I &= -\frac12 \pi \ln 2 \end{align*}
- \begin{align*} && \ln u &< u & \quad (u \geq 1)\\ \underbrace{\Rightarrow}_{u = \sqrt{x}} && \ln \sqrt{x} &< \sqrt{x} \\ \Rightarrow && \frac12 \ln x &< \sqrt{x} \\ \Rightarrow && \frac{\ln x}{x} &< \frac{2\sqrt{x}}{x} \\ &&&= \frac{2}{\sqrt{x}} \\ &&&\to 0 & (x \to \infty) \\ && x \ln x &= \frac{\ln 1/y}{y} \\ &&&= -\frac{\ln y}{y} \\ &&&\to 0 & (y \to \infty, x \to 0) \end{align*}
- \begin{align*} \int_{0}^{\frac{1}{2}\pi}x\cot x\,\mathrm{d}x &= \left [ x \ln(\sin x) \right]_0^{\pi/2} - \int_0^{\pi/2} \ln (\sin x) \d x \\ &= \left ( \frac{\pi}{2} \ln 1 - \lim_{x \to 0} x \ln (\sin x) \right) - \left ( -\frac12 \pi \ln 2 \right) \\ &= \frac12 \pi \ln 2 \end{align*}
1991 Paper 3 Q8
- The integral \(I_{k}\) is defined by \[ I_{k}=\int_{0}^{\theta}\cos^{k}x\,\cos kx\,\mathrm{d}x. \] Prove that \(2kI_{k}=kI_{k-1}+\cos^{k}\theta\sin k\theta.\)
- Prove that \[ 1+m\cos2\theta+\binom{m}{2}\cos4\theta+\cdots+\binom{m}{r}\cos2r\theta+\cdots+\cos2m\theta=2^{m}\cos^{m}\theta\cos m\theta. \]
- Using the results of (i) and (ii), show that \[ m\frac{\sin2\theta}{2}+\binom{m}{2}\frac{\sin4\theta}{4}+\cdots+\binom{m}{r}\frac{\sin2r\theta}{2r}+\cdots+\frac{\sin2m\theta}{2m} \] is equal to \[ \cos\theta\sin\theta+\cos^{2}\theta\sin2\theta+\cdots+\frac{1}{r}2^{r-1}\cos^{r}\theta\sin r\theta+\cdots+\frac{1}{m}2^{m-1}\cos^{m}\theta\sin m\theta. \]
Solution:
- \begin{align*} kI_k &= \int_0^\theta \cos^k x k\cos kx \d x \\ &= \left [\cos^k x \sin kx \right]_0^\theta - \int_0^\theta k \cos^{k-1} x \sin x \sin k x \d x \\ &= \cos^k \theta \sin k \theta - k\int_0^\theta \cos^{k-1} x \sin x \sin k x \d x \\ &= \cos^k \theta \sin k \theta + k\int_0^\theta \cos^{k-1} x (\cos (k-1) x - \cos k x \cos x) \d x \\ &= \cos^k\theta \sin k \theta + k I_{k-1} - kI_k \\ \end{align*} So \(2kI_k = kI_{k-1} + \cos^k \sin k \theta\)
- \begin{align*} && \sum_{r=0}^m \binom{m}{r} \cos 2r \theta &= \textrm{Re} \left ( \sum_{r=0}^m \binom{m}{r} \exp(i 2r \theta)\right) \\ &&&= \textrm{Re} \left ( \sum_{r=0}^m \binom{m}{r} \exp(i 2 \theta)^r\right) \\ &&&= \textrm{Re} \left ( \left (1+e^{2i \theta} \right)^m\right) \\ &&&= \textrm{Re} \left (\left (e^{i\theta}(e^{i \theta} +e^{-i\theta})\right)^m\right) \\ &&&= \textrm{Re} \left (\left (e^{i\theta}2\cos \theta\right)^m\right) \\ &&&= \textrm{Re} \left (2^m \cos^m \theta e^{im\theta}\right) \\ &&&= 2^m \cos^m \theta \cos m\theta \\ \end{align*}
- \begin{align*} \sum_{r=1}^m \binom{m}{r}\frac{\sin 2r \theta}{2r} &= \int_0^\theta \sum_{r=1}^m \binom{m}{r}\cos 2r x\d x \\ &= \int_0^\theta \left (2^m \cos^m x\cos mx- 1 \right ) \d x\\ &= 2^m I_m - \theta \\ &= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + 2^{m-1} I_{m-1} - \theta \\ &= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + \frac{2^{m-2}}{m-1}\cos^m \theta \sin m \theta+2^{m-2} I_{m-2} - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta + I_0 - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta + \int_0^\theta 1 \cdot 1 \d \theta - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta \\ &= \sum_{r=0}^{m} \frac{2^{r-1}}{r} \cos^{r} \theta \sin r \theta \end{align*}
1991 Paper 3 Q9
The parametric equations \(E_{1}\) and \(E_{2}\) define the same ellipse, in terms of the parameters \(\theta_{1}\) and \(\theta_{2}\), (though not referred to the same coordinate axes). \begin{alignat*}{2} E_{1}:\qquad & x=a\cos\theta_{1}, & \quad & y=b\sin\theta_{1},\\ E_{2}:\qquad & x=\dfrac{k\cos\theta_{2}}{1+e\cos\theta_{2}}, & \quad & y=\dfrac{k\sin\theta_{2}}{1+e\cos\theta_{2}}, \end{alignat*} where \(0< b< a,\) \(0< e< 1\) and \(0< k\). Find the position of the axes for \(E_{2}\) relative to the axes for \(E_{1}\) and show that \(k=a(1-e^{2})\) and \(b^{2}=a^{2}(1-e^{2}).\) {[}The standard polar equation of an ellipse is \(r=\dfrac{\ell}{1+e\cos\theta}.]\) By considering expressions for the length of the perimeter of the ellipse, or otherwise, prove that \[ \int_{0}^{\pi}\sqrt{1-e^{2}\cos^{2}\theta}\,\mathrm{d}\theta=\int_{0}^{\pi}\frac{1-e^{2}}{(1+e\cos\theta)^{2}}\sqrt{1+e^{2}+2e\cos\theta}\,\mathrm{d}\theta. \] Given that \(e\) is so small that \(e^{6}\) may be neglected, show that the value of either integral is \[ \tfrac{1}{64}\pi(64-16e^{2}-3e^{4}). \]
1991 Paper 3 Q10
The equation \[ x^{n}-qx^{n-1}+r=0, \] where \(n\geqslant5\) and \(q\) and \(r\) are real constants, has roots \(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}.\) The sum of the products of \(m\) distinct roots is denoted by \(\Sigma_{m}\) (so that, for example, \(\Sigma_{3}=\sum\alpha_{i}\alpha_{j}\alpha_{k}\) where the sum runs over the values of \(i,j\) and \(k\) with \(n\geqslant i>j>k\geqslant1\)). The sum of \(m\)th powers of the roots is denoted by \(S_{m}\) (so that, for example, \(S_{3}=\sum\limits_{i=1}^{n}\alpha_{i}^{3}\)). Prove that \(S_{p}=q^{p}\) for \(1\leqslant p\leqslant n-1.\) You may assume that for any \(n\)th degree equation and \(1\leqslant p\leqslant n\) \[ S_{p}-S_{p-1}\Sigma_{1}+S_{p-2}\Sigma_{2}-\cdots+(-1)^{p-1}S_{1}\Sigma_{p-1}+(-1)^{p}p\Sigma_{p}=0.] \] Find expressions for \(S_{n},\) \(S_{n+1}\) and \(S_{n+2}\) in terms of \(q,r\) and \(n\). Suggest an expression for \(S_{n+m},\) where \(m < n\), and prove its validity by induction.
Solution: Claim: \(S_p = q^p\) for \(1 \leq p \leq n-1\) Proof: When \(p = 1\), \(S_p = \Sigma_1 = q\) as expected. Note that \(\Sigma_i = 0\) for \(i = 2, \cdots, n-1\). Using \(S_p = S_{p-1}\Sigma_{1}-S_{p-2}\Sigma_{2}+\cdots+(-1)^{p-1+1}S_{1}\Sigma_{p-1}+(-1)^{p+1}p\Sigma_{p}\), we can see that \(S_p = qS_{p-q}\) when \(1 \leq p \leq n-1\), ie \(S_p = q^p\). Note that \begin{align*} S_n &= \sum \alpha_i^n \\ &= q\sum \alpha_i^{n-1} - \sum r \\ &= qS_{n-1} - nr \\ &= q^n - nr \\ \\ S_{n+1} &= \sum \alpha_i^{n+1} \\ &= q \sum \alpha_i^{n} - r \sum \alpha_i \\ &= q^{n+1} - rq \\ \\ S_{n+2} &= \sum \alpha_i^{n+2} \\ &= q \sum \alpha_i^{n+1} - r \sum \alpha_i^2 \\ &= q^{n+2} - rq^2 \\ \end{align*} Claim: \(S_{n+m} = q^{n+m} - rq^{m}\) Proof: The obvious