Problems

Solution:

1. \(A = \{1,4,6,7 \}, B = \{2,3,5,8 \}\)
2. \begin{align*} && \sum_{k=1}^m a_k^3 + \sum_{k=1}^m (c + b_k)^3 &= \sum_{k=1}^m a_k^3 + mc^3 + 3c^2 \sum_{k=1}^m b_k + 3c \sum_{k=1}^m b_k^2 + \sum_{k=1}^m b_k^3 \\ &&&= \sum_{k=1}^m a_k^3 + mc^3 + 3c^2 \sum_{k=1}^m a_k + 3c \sum_{k=1}^m a_k^2 + \sum_{k=1}^m b_k^3 \\ &&&= \sum_{k=1}^m a_k^3 + 3c \sum_{k=1}^m a_k^2+ 3c^2 \sum_{k=1}^m a_k + mc^3 + \sum_{k=1}^m b_k^3 \\ &&&= \sum_{k=1}^m (a_k+c)^3 + \sum_{k=1}^m b_k^3 \end{align*}
3. Note that \begin{align*} && \sum_{k=1}^m a_k^2 + \sum_{k=1}^m (c + b_k)^2 &= \sum_{k=1}^m a_k^2 + mc^2 + 2c \sum_{k=1}^m b_k +\sum_{k=1}^m b_k^2 \\ &&&= \sum_{k=1}^m (a_k+c)^2 + \sum_{k=1}^m b_k^2 \end{align*} Therefore if we take our balanced subsets of \(\{1,2,3,4,5,6,7,8\}\) and take \(A \cup (B+8)\) and \(B \cup (A+8)\) they will be balanced (including the cubes) so: \(A = \{1,4,6,7,10,11,13,16\}, B = \{2,3,5,8,9,12,14,15\}\)
4. Notice that: \begin{align*} \sum_{a \in A} (n+a)^2 &= |A|n^2 + 2n \sum_{a \in A} a + \sum_{a \in A} a^2 \\ &= 6 n^2 + 2n \sum_{b \in B}b + \sum_{b \in B} b^2 \\ &= n^2 + \sum_{b \in B}(n+b)^2 \\ \end{align*} Therefore consider the sets \(C = \{(n+i)^2 : i \in A\}, D = \{n^2\} \cup \{(n+i)^2 : i \in B\}\)

Solution:

1. \begin{align*} RHS &= \left(z - \frac{1}{z}\right)\left(z^m + \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right) \\ &= z^{m+1} + \frac{1}{z^{m-1}} - z^{m-1} - \frac{1}{z^{m+1}} + z^{m-1} - \frac{1}{z^{m-1}} \\ &= z^{m+1} - \frac{1}{z^{m+1}} \\ &= LHS \end{align*}. Claim: For \(n \geq 1\), $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ Proof: (By Induction) Base Case: (\(n=1\)). \begin{align*} LHS &= z^2 - \frac{1}{z^2} \\ &= (z-\frac1z)(z + \frac{1}{z}) \\ &= (z - \frac1z) \sum_{r=1}^1 \left ( z + \frac{1}{z} \right) \\ &= (z - \frac1z) \sum_{r=1}^1 \left ( z^{2r-1} + \frac{1}{z^{2r-1}} \right) \\ &= RHS \end{align*} as required. Inductive step: Suppose our result is true for some \(n=k\), then consider \(n = k+1\). \begin{align*} RHS &= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k+1} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) \\ &= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\ &= z^{2k} - \frac{1}{z^{2k}} + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\ &= z^{2k+2} - \frac{1}{z^{2k+2}} \\ &= LHS \end{align*}. Therefore if our result is true for \(n=k\) is true, it is true for \(n=k+1\). Since it is also true for \(n=1\) it is true for all \(n \geq 1\) but the principle of mathematical induction. Since \(\displaystyle z^{m+1} - \frac{1}{z^{m+1}} = \left(z + \frac{1}{z}\right)\left(z^m - \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right)\), we must have \(\displaystyle z^{2n}-\frac{1}{z^{2n}} = \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right)\)
2. 1. Since $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ we have \begin{align*} && e^{2n\theta i} - e^{-2n\theta i} &= \left(e^{\theta i} - e^{-\theta i}\right)\sum_{r=1}^n \left(e^{(2r-1)\theta i} + e^{-(2r-1)\theta i}\right) \\ \Rightarrow && 2i \sin 2n \theta &= 2i \sin \theta \sum_{r=1}^n 2 \cos (2r-1) \theta \\ \Rightarrow && \sin 2n \theta &= 2\sin \theta \sum_{r=1}^n \cos (2r-1) \theta \end{align*}
   2. When \(n = 2, \theta = \frac{\pi}{5}\) we have: \begin{align*} &&\sin \frac{4\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}) \\ &&\sin \frac{\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} - \cos \frac{2\pi}{5}) \\ &&\frac12 &= \cos \frac{\pi}{5} - \cos \frac{2 \pi}{5} \\ \Rightarrow && \cos \frac{2\pi}{5} &= \cos \frac{\pi}{5} - \frac12 \end{align*}
   3. When \(n = 7, \theta = \frac{\pi}{15}\) we have: \begin{align*} && \sin \frac{14 \pi}{15} &= 2 \sin \frac{\pi}{15} \sum_{r=1}^7 \cos (2r-1) \frac{\pi}{15} \\ \Rightarrow && \frac12 &= \cos \frac{\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}+ \cos \frac{7 \pi}{15}+ \cos \frac{9 \pi}{15}+ \cos \frac{11 \pi}{15}+ \cos \frac{13 \pi}{15} \\ &&&= -\cos \frac{16\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}- \cos \frac{8 \pi}{15}+ \cos \frac{9 \pi}{15}- \cos \frac{4 \pi}{15}- \cos \frac{2\pi}{15} \\ &&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \cos \frac{\pi}{5} + \cos \frac{\pi}{3} + \cos \frac{3 \pi}{5} \\ &&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \frac12 + \frac12 \\ \Rightarrow && \frac12 &= cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15} \end{align*}
3. By using \(z = e^{i \theta}\) we have that: \begin{align*} && z^{2n}-\frac{1}{z^{2n}} &= \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right ) \\ \Rightarrow && e^{2n \theta i} - e^{-2n \theta i} &= (e^{\theta i} + e^{-\theta i}) \sum_{r=1}^n (e^{(2r-1)\theta i} - e^{(2r-1) \theta i}) \\ \Rightarrow && 2i \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n 2i \sin(2r-1) \theta \\ \Rightarrow && \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n \sin(2r-1) \theta \end{align*} When \(n = 3, \theta = \frac{\pi}{14}\) we must have: \begin{align*} &&\sin \frac{3 \pi}{7} &= 2 \cos \frac{\pi}{14}( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ &&&= 2 \sin \left (\frac{\pi}{2} - \frac{\pi}{14} \right)( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ &&&= 2 \sin \frac{3\pi}{7} ( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ \Rightarrow && \frac12 &= \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14} \end{align*} as required.

Solution:

1. \(\,\) \begin{align*} && \frac1{r+1} - \frac1r + \frac1{r^2} &= \frac{-1}{r(r+1)} + \frac{1}{r^2} \\ &&&= \frac{r+1-r}{r^2(r+1)} \\ &&&= \frac{1}{r^2(r+1)} \end{align*} Therefore \begin{align*} && \sum_{r=1}^N \frac1{r^2(r+1)} &= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r + \frac1{r^2} \right) \\ &&&= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r\right) + \sum_{r=1}^N \frac1{r^2} \\ &&&=\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \\ \end{align*} therefore \begin{align*} && \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} &= \lim_{N \to \infty } \sum_{r=1}^{N} \frac{1}{r^2(r+1)} \\ &&&= \lim_{N \to \infty } \left (\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \right) \\ &&&= -1 +\lim_{N \to \infty } \sum_{r=1}^N \frac1{r^2} \\ &&&= -1 + \sum_{r=1}^\infty \frac1{r^2} \\ &&&= \frac{\pi^2}{6}-1 \end{align*}
2. Note that \begin{align*} && \frac{1}{r^2(r+1)(r+2)} &= \frac{Ar+B}{r^2} + \frac{C}{r+1} + \frac{D}{r+2} \\ &&&= \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r} \end{align*} So \begin{align*} && \sum_{r=1}^N \frac{1}{r^2(r+1)(r+2)} &= \sum_{r=1}^N \left ( \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r} \right ) \\ &&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} + \frac{1}{2} - \frac14 \cdot \frac1{3} - \frac34 \frac11 + \\ &&& \quad \quad \quad + \frac13 - \frac14\frac14 - \frac34\frac12 + \\ &&& \quad \quad \quad + \frac14 - \frac14\frac15 - \frac34\frac13 + \\ &&&\quad \quad \quad+ \cdots + \\ &&&\quad \quad \quad + \frac1{N+1} - \frac14\frac1{N+2} - \frac34\frac1N \\ &&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1} \\ \\ \Rightarrow && \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)} &= \lim_{N \to \infty} \left [ \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1}\right] \\ &&&= \frac{\pi^2}{12} -\frac58 \end{align*}
3. Notice that \(\frac{1}{r^2(r+1)^2} - \frac{2}{r^2(r+1)} = \frac{1-2(r+1)}{r^2(r+1)^2} = \frac{-1-2r}{r^2(r+1)^2} = \frac{1}{(r+1)^2} - \frac{1}{r^2}\) and so \begin{align*} && \sum_{r=1}^N \frac{1}{r^2(r+1)^2} &= \sum_{r=1}^N \left ( \frac{2}{r^2(r+1)} +\frac{1}{(r+1)^2} - \frac{1}{r^2} \right) \\ &&&= \sum_{r=1}^N \frac{2}{r^2(r+1)} +\frac{1}{(N+1)^2} - 1 \\ \end{align*} and the result follows as \(N \to \infty\)

Solution:

1. \(\,\)
   

   + - ↺
   Notice that \begin{align*} && (a+b)^2 &= PQ^2 + (a-b)^2 \\ \Rightarrow && PQ^2 &= 4ab \\ && (b+c)^2 &= QR^2 + (c-b)^2 \\ \Rightarrow && QR^2 &= 4bc \\ && (a+c)^2 &= PR^2 + (a-c)^2 \\ \Rightarrow && PR^2 &= 4ac \\ \Rightarrow && 2\sqrt{ac} &= 2\sqrt{ab}+2\sqrt{bc} \\ \Rightarrow && \frac{1}{\sqrt{b}} &= \frac{1}{\sqrt{c}} + \frac1{\sqrt{a}} \\ \end{align*} Let \(x, y, z = \frac{1}{\sqrt{a}}, \frac1{\sqrt{b}}, \frac{1}{\sqrt{z}}\) so we would like to prove that \(2(x^4+y^4+z^4) = (x^2+y^2+z^2)^2\) or \(x^4+y^4+z^4 = 2x^2y^2+2y^2z^2+2z^2x^2\). We also have \begin{align*} && y &= x+z \\ \Rightarrow &&y^2 &= x^2+z^2+2xz \\ \Rightarrow && (y^2-x^2-z^2)^2 &= 4x^2z^2 \\ \Rightarrow && y^4+x^4+z^4 - 2x^2y^2-2y^2z^2+2x^2z^2 &= 4x^2z^2\\ \Rightarrow && y^4+x^4+z^4 &= 2x^2y^2+2y^2z^2+2z^2x^2 \end{align*}
2. Notice that subject to \(y > z > x\) all these steps are reversible, so we must have the equality we desire

Solution:

1. Claim: For all \(n \geq 1\) \(T_{2n} = A_{2n} + B_{2n}\sqrt{a(a+1)}\) where \(A_{2n}, B_{2n}\) are integers and \(A_{2n}^2 = a(a+1)B_{2n}^2+1\) Proof: (By induction) Base case: \(n =1\). \begin{align*} && T_2 &= (\sqrt{a+1}+\sqrt{a})^2 \\ &&&= a+1+2\sqrt{a(a+1)}+a \\ &&&=2a+1+2\sqrt{a(a+1)} \\ \Rightarrow && A_2 &= 2a+1 \\ && B_2 &= 2 \\ && A_2^2 &= 4a^2+4a+1 \\ && a(a+1)B_1^2 + 1 &= 4a^2+4a+1 \end{align*} Therefore our base case is true. Suppose it is true for some \(n = k\) then consider \(n = k+1\) we must have \(T_{2k} = A_{2k}+B_{2k}\sqrt{a(a+1)}\) \begin{align*} && T_{2(k+1)} &= T_{2k} (\sqrt{a+1}+\sqrt{a})^2 \\ &&&= \left (A_{2k}+B_{2k}\sqrt{a(a+1)} \right)\left (2a+1+2\sqrt{a(a+1)} \right) \\ &&&= (2a+1)A_{2k}+2a(a+1)B_{2k} + (2A_{2k}+(2a+1)B_{2k})\sqrt{a(a+1)} \\ \Rightarrow && A_{2(k+1)} &= (2a+1)A_{2k}+2a(a+1)B_{2k} \in \mathbb{Z} \\ && B_{2(k+1)} &= 2A_{2k}+(2a+1)B_{2k} \in \mathbb{Z} \\ && A_{2(k+1)}^2 &= \left ( (2a+1)A_{2k}+2a(a+1)B_{2k} \right)^2 \\ &&&= (2a+1)^2A_{2k}^2+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k} \\ &&&= a^2(a+1)^2(2a+1)^2B_{2k}^2+(2a+1)^2\\ &&&\quad\quad+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k}\\ && a(a+1)B_{2(k+1)}^2 + 1 &= a(a+1)\left ( 2A_{2k}+(2a+1)B_{2k} \right)^2 \\ &&&= 4a(a+1)A_{2k}^2 + 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1\\ &&&= 4a^2(a+1)^2B_{2k}^2+4a(a+1) + \\ &&&\quad\quad+ 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1 \end{align*} So our relation holds ad therefore by induction we're done.
2. When \(n\) is odd, notice that \begin{align*} && T_n &= (\sqrt{a+1}+\sqrt{a})(A_m+B_m\sqrt{a(a+1)})\\ &&&= \sqrt{a+1}(\underbrace{A_m+aB_m}_{C_n})+\sqrt{a}(\underbrace{(a+1)B_m+A_m}_{D_n}) \\ \\ && (a+1)C_n^2 &= (a+1)(A_m+aB_m)^2 \\ &&&= (a+1)(A_m^2+2aA_mB_m+B_m^2) \\ &&&= (a+1)(a(a+1)B_m^2+1+2aA_mB_m+B_m^2) \\ &&&= a(a+1)^2B_m^2 + 2a(a+1)A_mB_m+(a+1)B_m^2+a+1 \\ && aD_n^2+1 &= a((a+1)^2B_m + 2(a+1)A_mB_m + A_m^2)+1 \\ &&&= a(a+1)^2B_m + 2a(a+1)A_mB_m + aB_{m}^2+a+1 \end{align*}
3. For even \(n\) \(T_n = \sqrt{a(a+1)B_{2n}^2+1} + \sqrt{a(a+1)B_{2n}^2}\) For odd \(n\), \(T_n = \sqrt{aD_n^2+1}+ \sqrt{aD_n^2}\) therefore it is always the sum of the square root of two consecutive integers.

Solution: The complex numbers represent an equilateral triangle iff \(\gamma\) is a \(\pm 60^\circ\) rotation of \(\beta\) around \(\alpha\), ie \begin{align*} && \gamma - \alpha &= \omega(\beta - \alpha) \\ \Leftrightarrow && \omega &= \frac{\gamma - \alpha}{\beta - \alpha} \\ \Leftrightarrow && -1 &= \left (\frac{\gamma - \alpha}{\beta - \alpha} \right)^3 \\ \Leftrightarrow && -(\beta - \alpha)^3 &=(\gamma - \alpha)^3 \\ \Leftrightarrow && 0 &= (\gamma-\alpha)^3+(\beta-\alpha)^3 \\ &&&= \gamma^3-3\gamma^2\alpha +3\gamma\alpha^2-\alpha^3 +\beta^3-3\beta^2\alpha+3\beta\alpha^2-\alpha^3 \\ &&&= (\beta + \gamma - 2\alpha)(\alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta) \\ \Leftrightarrow && 0 &= \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta \end{align*} The roots of the equation \(z^3+az^2+bz+c = 0\) represents the vertices of an equilateral triangle iff \(a^2-3b = (\alpha+\beta+\gamma^2) - 3(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta = 0\) as erquired. Suppose \(a^2 = 3b\), then consider \(z = pw +q\), we must have \begin{align*} && 0 &= (pw+q)^3+a(pw+q)^2 + b(pw+q)+c \\ &&&= p^3w^3 +(3p^2q+ap^2)w^2+(3pq^2+2apq+bp)w+(q^3+aq^2+bq+c) \\ &&&= p^3w^3+p^2(3q+a)w^2+p(3q^2+2aq+b)w+(q^3+aq^2+bq+c) \\ \end{align*} We need to check if \(\left(\frac{3q+a}{p} \right)^2 = 3 \left (\frac{3q^2+2qa+b}{p^2} \right)\). Clearly the denominators match, so consider the numerators \begin{align*} && (3q+a)^2 &= 9q^2+6aq+a^2 \\ &&&= 9q^2+6aq+3b \\ &&&= 3(3q^2+2qa+b) \end{align*} as required

Solution: \begin{align*} && \tan \left (\arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) \right) &= \frac{\frac1{a+b}+\frac1{a+c}}{1-\frac{1}{(a+b)(a+c)}} \\ &&&= \frac{a+c+a+b}{(a+b)(a+c)-1} \\ &&&= \frac{2a+b+c}{a^2+ab+ac+bc-1} \\ &&&= \frac{2a+b+c}{2a^2+ab+ac} \\ &&&= \frac{1}{a} \\ &&&= \tan \arctan \frac1a\\ \Rightarrow && \arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) &= \arctan \frac{1}{a} + n \pi \end{align*} Since \(\arctan x \in (-\frac{\pi}{2}, \frac{\pi}{2})\) the LHS \(\in (0, \pi)\) so \(n = 0\). \begin{align*} a=p+q, b = s, c = t:&& \arctan \! \!\left(\!\frac1 {p+q+s}\!\right) + \arctan \! \!\left(\!\frac 1{p+q+t}\!\right) &= \arctan \left ( \frac{1}{p+q} \right) \\ a=p+r, b= u, c = v && \arctan \! \!\left(\!\frac 1 {p+r+u}\!\right) + \arctan \! \!\left(\!\frac1 {p+r+v}\!\right) &= \arctan \! \!\left(\!\frac1 {p+r}\!\right) \\ a = p, b = q, c = r:&& \arctan \left ( \frac{1}{p+q} \right) +\arctan \! \!\left(\!\frac1 {p+r}\!\right) &= \arctan \left ( \frac1p \right) \end{align*} and the result follows. Taking \(p = 7\) we need to solve \[ \begin{cases} q+s &= 6 \\ q+t &= 14 \\ r+u &= 75 \\ r+v &= 180 \end{cases} \] also satisfying \(qr = 50\) etc, so say \(q = 1, r = 50, s = 5, v=25\)

Solution: To obtain the coefficient of \(x^n\) on the RHS we clearly have \(\displaystyle \binom{2n}n\). To obtain the coefficient of \(x^n\) on the LHS we can obtain \(x^s\) from the first bracket and \(x^{n-s}\) from the second bracket, ie \(\displaystyle \sum_{s=0}^n \binom{n}{s}\binom{n}{n-s} = \sum_{s=0}^n \binom{n}{s}\binom{n}{s} = \sum_{s=0}^n \binom{n}{s}^2\)

1. Consider \((1-x)^n(1+x)^n = (1-x^2)^n\), then the coefficient of \(x^n\) (if \(n\) is even) is for the RHS \(\displaystyle (-1)^{n/2} \binom{n}{n/2}\). For the LHS, we can obtain \(x^n\) via \(x^s\) and \(x^{n-s}\) which is \(\displaystyle \sum_{s=0}^n (-1)^s\binom{n}{s}\binom{n}{n-s} = \sum_{s=0}^n (-1)^s\binom{n}{s}^2\)
2. Notice that \begin{align*} && \sum_{t=1}^ n 2t { n \choose t}^2 &= n {2n\choose n} \\ \Leftrightarrow && \sum_{t=1}^ n 2t \frac{n}{t} { n-1 \choose t-1}\binom{n}{t} &= n \frac{2n}{n}{2n-1\choose n-1} \\ \Leftrightarrow && \sum_{t=1}^ n { n-1 \choose t-1}\binom{n}{t} &= {2n-1\choose n-1} \\ \Leftrightarrow && \sum_{t=1}^ n { n-1 \choose t-1}\binom{n}{n-t} &= {2n-1\choose n-1} \\ \end{align*} but this is exactly what we would obtain by considering the coefficient of \(x^{n-1}\) in \((1+x)^{n-1}(1+x)^n \equiv (1+x)^{2n-1}\)

Solution: Claim: \(u^2_{n+2} + u^2_{n-1} = 2( u^2_{n+1} + u^2_n )\) Proof: (By Induction). (Base Case): \(n = 1\) \begin{align*} && LHS &= u_{n+2}^2 + u_{n-1}^2 \\ &&&= u_3^2 + u_0^2 \\ &&&= 3^2 + 1^2 = 10\\ && RHS &= 2(u_{n+1}^2+u_n^2) \\ &&&= 2(2^2 + 1^2) \\ &&&= 10 \end{align*} Therefore the base case is true. (Inductive hypothesis) Suppose \(u^2_{n+2} + u^2_{n-1} = 2( u^2_{n+1} + u^2_n )\) is true for some \(n = k\), ie \(u^2_{k+2} + u^2_{k-1} = 2( u^2_{k+1} + u^2_k )\), the consider \(n = k+1\) \begin{align*} && LHS &= u_{k+1+2}^2 + u_{k+1-1}^2 \\ &&&= (u_{k+1}+u_{k+2})^2+u_k^2 \\ &&&= u_{k+2}^2+u_{k+1}^2+ u_k^2 + 2u_{k+1}u_{k+2} \\ &&&= u_{k+2}^2+u_{k+1}^2+ u_k^2 + 2u_{k+1}(u_{k+1}+u_k) \\ &&&= u_{k+2}^2 + u_{k+1}^2+u_k^2+2u_{k+1}^2+2u_{k+1}u_k \\ &&&= u_{k+1}^2+2u_{k+1}^2+ u_{k+1}^2+u_k^2+2u_{k+1}u_k \\ &&&= u_{k+2}^2+2u_{k+1}^2+ (u_{k+1}+u_k)^2 \\ &&&= u_{k+2}^2+2u_{k+1}^2+ u_{k+2}^2 \\ &&&=2(u_{k+2}^2+u_{k+1}^2) \\ &&&= RHS \end{align*} Therefore it is true for \(n = k+1\). Therefore by the principle of mathematical induction it is true for all \(n \geq 1\) Claim: $ u_{2n} = u^2_n + u^2_{n-1} \quad \mbox{ and }\quad u_{2n+1} = u^2_{n+1} - u^2_{n-1} $ Proof: Notice that \(\begin{pmatrix}u_{n+1} \\ u_n \end{pmatrix}= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix}1 \\1 \end{pmatrix}\), in particular \begin{align*} && \begin{pmatrix}u_{n}& u_{n-1} \\ u_{n-1} & u_{n-2} \end{pmatrix}&= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} \\ \Rightarrow && \begin{pmatrix}u_{2n}& u_{2n-1} \\ u_{2n-1} & u_{2n-2} \end{pmatrix}&= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{2n} \\ &&&= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} \\ &&&=\begin{pmatrix}u_{n}& u_{n-1} \\ u_{n-1} & u_{n-2} \end{pmatrix}\begin{pmatrix}u_{n}& u_{n-1} \\ u_{n-1} & u_{n-2} \end{pmatrix}\\ &&&= \begin{pmatrix}u_{n}^2+u_{n-1}^2& u_{n-1}(u_n+u_{n-2}) \\ u_{n-1}(u_n+u_{n-2}) & u_{n-1}^2+u_{n-2}^2 \end{pmatrix} \end{align*} Therefore \(u_{2n} = u_{n}^2+u_{n-1}^2\) and \(u_{2n+1} = u_n(u_{n+1}+u_{n-1}) =(u_{n+1}-u_{n-1})(u_{n+1}-u_{n-1}) = u_{n+1}^2-u_{n-1}^2\)

Solution: Plugging in \(x = i\) we obtain \((1-i)^4 = (-2i)^2 = -4 \Rightarrow b = -4\). \begin{align*} x^4(1-x)^4 &= x^4(1-4x+6x^2-4x^3+x^4) \\ &= x^8-4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) - x^6 -4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) -4x^5(x^2+1)+4x^5 +5x^6-4x^5+x^4 \\ &= (x^6-4x^5)(x^2+1) +5x^4(x^2+1)-5x^4+x^4 \\ &= (x^6-4x^5+5x^4)(x^2+1) -4x^2(x^2+1)+4x^2 \\ &= (x^6-4x^5+5x^4-4x^2)(x^2+1) +4(x^2+1)-4 \\ &= (x^6-4x^5+5x^4-4x^2+4)(x^2+1) -4 \\ \end{align*} So \begin{align*} \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \d x &= \int_0^1 (x^6-4x^5+5x^4-4x^2+4) - \frac{4}{1+x^2} \d x \\ &= \frac17 - \frac46+1-\frac43+4 - \pi \\ &= \frac{22}7 - \pi \end{align*} \begin{align*} \int_0^1 x^4(1-x)^4 \d x &= B(5,5) \\ &= \frac{4!4!}{9!} \\ &= \frac1{630} \end{align*} Therefore since \(0 < \frac{x^4(1-x)^4}{1+x^2} < x^4(1-x)^4\) we must have that \begin{align*} && 0 &< \frac{22}7 - \pi \\ \Rightarrow && \pi & < \frac{22}{7} \\ && \frac{22}{7} - \pi &< \frac1{630} \\ \Rightarrow && \frac{22}{7} - \frac1{630} &< \pi \end{align*} which is what we wanted.

Solution: \begin{align*} \frac{1}{(1-ax)(1-bx)} &=\frac{1}{b-a} \l \frac{b}{1-bx}-\frac{a}{1-ax}\r \\ &= \frac{1}{b-a} \l \sum_{k=0}^{\infty} b(bx)^k-\sum_{k=0}^{\infty} a(ax)^k \r \\ &= \frac{1}{b-a} \sum_{k=0}^{\infty} \l b^{k+1} - a^{k+1} \r x^k \end{align*} Therefore \(\displaystyle c_m = \frac{b^{k+1}-a^{k+1}}{b-a}\). \begin{align*} c_m^2 &= \frac{(b^{m+1}-a^{m+1})^2}{(b-a)^2} \\ &= \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \end{align*} \begin{align*} \sum_{m=0}^{\infty} c_m x^m &= \sum_{m=0}^{\infty} \l \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \r x^m \\ &= \frac{1}{(b-a)^2} \l \sum_{m=0}^{\infty} a^{2m+2} x^m-2\sum_{m=0}^{\infty} (ab)^{m+1} x^m+\sum_{m=0}^{\infty} b^{2m+2} x^m \r \\ &= \frac{1}{(b-a)^2} \l a^2\sum_{m=0}^{\infty} a^{2m} x^m-2ab\sum_{m=0}^{\infty} (ab)^{m} x^m+b^2\sum_{m=0}^{\infty} b^{2m} x^m \r \\ &= \frac{1}{(b-a)^2} \l \frac{a^2}{1-a^2x^2} - \frac{2ab}{1-abx} + \frac{b^2}{1-b^2x^2}\r \\ &= \frac{1+ab}{(1-a^2x)(1-abx)(1-b^2x)} \end{align*} Where geometric series will converge if \(|a^2x| < 1, |b^2x| < 1, |abx| < 1\), ie \(|x| < \min (\frac{1}{a^2}, \frac{1}{b^2} )\)

Solution:

1. \begin{align*} \left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2} &= x^2-2ax+a^2 + 4x^2 -4bx+b^2 + 9x^2-6cx + c^2 \\ &= (1^2 + 2^2 + 3^2)x^2 - 2x(a+2b+3c) +a^2+b^2 + c^2 \\ &= 14x^2 - 2x(7x) + a^2 + b^2 + c^2 \\ &= a^2 + b^2 + c^2 \end{align*}
2. \begin{align*} \left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2} &= (2^2+3^2+3^2)x^2 - 2x(2a+3b+3c) + (a^2 + b^2+c^2) \\ &= 22x^2 - 2x(11x) + a^2+b^2+c^2 \\ &= a^2+b^2+c^2 \end{align*}