Edexcel Core Pure 1

Complete the square and remember that $i^2=-1$.

Completing the square gives $(z-2)^2+9=0$, so $(z-2)^2=-9$.

Therefore $z-2=\pm3i$, giving the conjugate roots $2+3i$ and $2-3i$.

Answer: $z=2+3i$ or $z=2-3i$.

Calculate the determinant before applying the inverse formula for a two-by-two matrix.

The determinant is $2(1)-1(1)=1$, so $M^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$.

The equations are $M\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}7\\4\end{pmatrix}$. Premultiplying by $M^{-1}$ gives $\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}7-4\\-7+8\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}$.

Answer: $M^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$, $x=3$, $y=1$.

After assuming the formula for $n=k$, add the next term $(k+1)(k+2)$ and factorise.

For $n=1$, the left side is $1(2)=2$ and the right side is $1(2)(3)/3=2$, so the statement is true.

Assume it is true for a positive integer $k$: $\sum_{r=1}^{k}r(r+1)=k(k+1)(k+2)/3$.

Then $$\begin{aligned}\sum_{r=1}^{k+1}r(r+1)&=\frac{k(k+1)(k+2)}3\\&\quad +(k+1)(k+2)\\&=\frac{(k+1)(k+2)(k+3)}3.\end{aligned}$$

This is the required formula with $n=k+1$. The statement holds at $n=1$, and truth at any positive integer implies truth at the next; therefore it holds for every positive integer by induction.

Answer: $\displaystyle\sum_{r=1}^{n}r(r+1)=\dfrac{n(n+1)(n+2)}3$ for all positive integers $n$.

$AB=\begin{pmatrix}2&1\\0&1\end{pmatrix}$, whereas $BA=\begin{pmatrix}2&2\\0&1\end{pmatrix}$.

The top-right entries differ, so $AB\neq BA$. Matrix multiplication is not commutative in general; factors cannot be reordered without justification.