**(a)** $y = 2\sqrt{a} \, x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \sqrt{a} \, x^{-\frac{1}{2}}$ | M1 | $\frac{dy}{dx} = \pm k x^{-\frac{1}{2}}$ or $k\frac{dy}{dx} = c$ or their $\frac{dy}{dx}$ or their $\frac{dx}{dy}$

or (implicitly) $2y \frac{dy}{dx} = 4a$ | | 

or (chain rule) $\frac{dy}{dx} = 2a \times \frac{1}{2\sqrt{at}}$ | | 

When $x = at^2$: $\frac{dy}{dx} = \frac{\sqrt{a}}{\sqrt{at^2}} = \frac{\sqrt{a}}{\sqrt{at}} = \frac{1}{t}$ | A1 | $\frac{dy}{dx} = \frac{1}{t}$

or $\frac{dy}{dx} = \frac{4a}{2(2at)} = \frac{1}{t}$ | | 

$T: \quad y - 2at = \frac{1}{t}(x - at^2)$ | M1 | Applies $y - 2at = \text{their } m_T \left(x - at^2\right)$. Their $m_T$ must be a function of $t$ from calculus.

$T: \quad ty - 2at^2 = x - at^2$ | | 

$T: \quad ty = x + at^2$ | A1 cso | Correct solution. | [4]

**(b)** At $Q$, $x = 0 \Rightarrow y = \frac{at^2}{t} = at \Rightarrow Q(0, at)$ | B1 | $y = at$ or $Q(0, at)$ | [1]

**(c)** $S(a, 0)$ | | 

$m(PQ) = \frac{at - 2at^2}{0 - at^2} = \frac{-at}{-at^2} = \frac{1}{t}$ | M1 | A correct method for finding either $m(PQ)$ or $m(SQ)$ for their $Q$ or $S$.

$m(SQ) = \frac{at - 0}{0 - a} = \frac{at}{-a} = -t$ | A1 | $m(PQ) = \frac{1}{t}$ and $m(SQ) = -t$

$m(PQ) \times m(SQ) = \frac{1}{t} \times -t = -1 \Rightarrow PQ \perp SQ$ | A1 cso | Shows $m(PQ) \times m(SQ) = -1$ and conclusion. | [3]

**Total: [8]**

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## Question 8(a):

$\sum_{r=1}^{n} r(2r-1) = \frac{1}{6}n(n+1)(4n-1)$

$n = 1$: $\text{LHS} = \sum_{r=1}^{1} r(2r-1) = 1$ | B1 | $\frac{1}{6}(1)(2)(3) = 1$ seen

$\text{RHS} = \frac{1}{6}(1)(2)(3) = 1$

As LHS = RHS, the summation formula is true for $n = 1$.

Assume that the summation formula is true for $n = k$.

ie. $\sum_{r=1}^{k} r(2r-1) = \frac{1}{6}k(k+1)(4k-1)$.

With $n = k+1$ terms the summation formula becomes:

$\sum_{r=1}^{k+1} r(2r-1) = \frac{1}{6}k(k+1)(4k-1) + (k+1)(2(k+1)-1)$ | M1 | $S_{k+1} = S_k + u_{k+1}$ with $S_k = \frac{1}{6}k(k+1)(4k-1)$.

$= \frac{1}{6}k(k+1)(4k-1) + (k+1)(2k+1)$ | | 

$= \frac{1}{6}(k+1)(k(4k-1) + 6(2k+1))$ | dM1 | Factorise by $\frac{1}{6}(k+1)$

$= \frac{1}{6}(k+1)(4k^2 + 11k + 6)$ | A1 | $(4k^2 + 11k + 6)$ or equivalent quadratic seen

$= \frac{1}{6}(k+1)(k+2)(4k+3)$ | | 

$= \frac{1}{6}(k+1)((k+1)+1)(4(k+1)-1)$ | dM1 | Correct completion to $S_{k+1}$ in terms of $k+1$ dependent on both Ms.

If the summation formula is true for $n = k$, then it is shown to be true for $n = k+1$. As the result is true for $n = 1$, it is now also true for all $n$ and $n \in \mathbb{Z}^+$ by mathematical induction. | A1 cso | Conclusion with all 4 underlined elements that can be seen anywhere in the solution | [6]

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## Question 8(b):

$\sum_{r=n+1}^{3m} r(2r-1) = S_{3m} - S_n$ | | 

$= \frac{1}{6}3m(3m+1)(12m-1) - \frac{1}{6}n(n+1)(4n-1)$ | M1 | Use of $S_{3m} - S_n$ or $S_{3m} - S_{n+1}$ with the result from (a) used at least once. Correct un-simplified expression.

$= \frac{1}{6}n\{3(3n+1)(12n-1) - (n+1)(4n-1)\}$ | A1 | 

$= \frac{1}{6}n\{3(36n^2 + 9n - 1) - (4n^2 + 3n - 1)\}$ | dM1 | Factorises out $\frac{1}{n}$ or $\frac{1}{3}n$ and an attempt to open up the brackets.

$= \frac{1}{6}n\{108n^2 + 27n - 3 - 4n^2 - 3n + 1\}$ | | 

$= \frac{1}{6}n\{104n^2 + 24n - 2\}$ | | 

$= \frac{1}{3}n(52n^2 + 12n - 1)$ | A1 | $= \frac{1}{3}n(52n^2 + 12n - 1)$

$\{a = 52, b = 12, c = -1\}$ | | [4]

**Total: [10]**

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