2 problems found
Given that \[ 5x^{2}+2y^{2}-6xy+4x-4y\equiv a\left(x-y+2\right)^{2} +b\left(cx+y\right)^{2}+d\,, \] find the values of the constants \(a\), \(b\), \(c\) and \(d\). Solve the simultaneous equations \begin{align*} 5x^{2}+2y^{2}-6xy+4x-4y&=9\,, \\ 6x^{2}+3y^{2}-8xy+8x-8y&=14\,. \end{align*}
Solution: \(a\left(x-y+2\right)^{2} +b\left(cx+y\right)^{2}+d\, \equiv (a + bc^2)x^2 + (a+b)y^2 + (-2a+2bc)xy + (4a)x+(-4ay) + 4a+d\) so we want to solve \[ \begin{cases} a + bc^2 &= 5 \\ a+b &= 2 \\ 2bc - 2a &= -6 \\ 4a &= 4 \\ -4a &= 4 \\ 4a+d &= -9 \end{cases} \Rightarrow a = 1, b = 1, c = -2, d = -13 \] Therefore we have: \((x-y+2)^2 + (2x+y)^2-13 = 0\) and our simultaneous equations will be: \[ \begin{cases} (x-y+2)^2 + (-2x+y)^2 &= 13 \\ 2(x-y+2)^2 + (-2x+y)^2 &= 22 \end{cases} \] which are simultaneous equations in \((x-y+2)^2\) and \((-2x+y)^2\) which solve to \((x-y+2)^2 = 9, (-2x+y)^2 = 4 \) so we need to solve \(4\) sets of simultaneous equations: \begin{align*} &\begin{cases} x - y + 2 &= 3 \\ -2x + y &= 2 \end{cases} &&\Rightarrow (x,y) = (-3, -4) \\ &\begin{cases} x - y + 2 &= -3 \\ -2x + y &= 2 \end{cases} &&\Rightarrow (x,y) = (3, 8) \\ &\begin{cases} x - y + 2 &= 3 \\ -2x + y &= -2 \end{cases} &&\Rightarrow (x,y) = (1, 0) \\ &\begin{cases} x - y + 2 &= -3 \\ -2x + y &= -2 \end{cases} &&\Rightarrow (x,y) = (7, 12) \\ \end{align*} So \((x,y) = (-3, -4), (3, 8), (1, 0), (7,12)\)
By considering the coefficient of \(x^{n}\) in the identity \((1-x)^{n}(1+x)^{n}=(1-x^{2})^{n},\) or otherwise, simplify \[ \binom{n}{0}^{2}-\binom{n}{1}^{2}+\binom{n}{2}^{2}-\binom{n}{3}^{2}+\cdots+(-1)^{n}\binom{n}{n}^{2} \] in the cases (i) when \(n\) is even, (ii) when \(n\) is odd.
Solution: The coefficient of \(x^n\) on the LHS is \begin{align*} && (1-x^2)^n &= (1-x)^n(1+x)^n \\ [x^n]: && \begin{cases} (-1)^{\lfloor \frac{n}2 \rfloor}\binom{n}{\lfloor \frac{n}2 \rfloor} &\text{if } n\text{ even} \\ 0 & \text{otherwise} \end{cases} &= \sum_{i=0}^n \underbrace{(-1)^i\binom{n}{i}}_{\text{take }(-x)^i\text{ from first bracket}} \cdot \underbrace{\binom{n}{n-i}}_{\text{take }x^{n-i}\text{ from second bracket}} \\ &&&= \sum_{i=0}^n (-1)^i\binom{n}{i}\binom{n}{i} \\ &&&= \sum_{i=0}^n (-1)^i\binom{n}{i}^2\\ \end{align*}