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IAL 2024 Jan Q3

A Level / Edexcel / FP2

IAL 2024 Jan Paper · Question 3

(a) Show that for r⩾1r \geqslant 1

rr(r+1)+r(r−1)= A(r(r+1)−r(r−1))\begin{align*} \frac{r}{\sqrt{r(r + 1)} + \sqrt{r(r - 1)}} =\,& A\left(\sqrt{r(r + 1)} - \sqrt{r(r - 1)}\right) \end{align*}

where AA is a constant to be determined.

(2)

(b) Hence use the method of differences to determine a simplified expression for

∑r=1nrr(r+1)+r(r−1)\begin{align*} \sum_{r=1}^{n} \frac{r}{\sqrt{r(r + 1)} + \sqrt{r(r - 1)}} \end{align*}
(3)

(c) Determine, as a surd in simplest form, the constant kk such that

∑r=1nkrr(r+1)+r(r−1)=∑r=1nr\begin{align*} \sum_{r=1}^{n} \frac{kr}{\sqrt{r(r + 1)} + \sqrt{r(r - 1)}} = \sqrt{\sum_{r=1}^{n} r} \end{align*}
(2)
解答

(a)

To find AA, we rationalise the denominator by multiplying the numerator and denominator by the conjugate r(r+1)−r(r−1)\sqrt{r(r+1)} - \sqrt{r(r-1)}:

 rr(r+1)+r(r−1)= r(r(r+1)−r(r−1))(r(r+1)+r(r−1))(r(r+1)−r(r−1))= r(r(r+1)−r(r−1))r(r+1)−r(r−1)= r(r(r+1)−r(r−1))r2+r−r2+r= r(r(r+1)−r(r−1))2r= 12(r(r+1)−r(r−1))\begin{align*} &\, \frac{r}{\sqrt{r(r+1)}+\sqrt{r(r-1)}}\\[4mm] = &\, \frac{r\left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right)}{\left(\sqrt{r(r+1)}+\sqrt{r(r-1)}\right)\left(\sqrt{r(r+1)}-\sqrt{r(r-1)}\right)}\\[4mm] = &\, \frac{r\left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right)}{r(r+1) - r(r-1)}\\[4mm] = &\, \frac{r\left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right)}{r^2 + r - r^2 + r}\\[4mm] = &\, \frac{r\left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right)}{2r}\\[4mm] = &\, \frac{1}{2}\left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right) \end{align*}

Thus, comparing to the given form, A=12A = \frac{1}{2}.

(b)

Using the result from (a), we can rewrite the sum:

∑r=1nrr(r+1)+r(r−1)= 12∑r=1n(r(r+1)−r(r−1))\begin{align*} \sum_{r=1}^{n}\frac{r}{\sqrt{r(r+1)}+\sqrt{r(r-1)}} = &\, \frac{1}{2} \sum_{r=1}^{n} \left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right) \end{align*}

Writing out the terms to show the method of differences:

r=1: 12(2−0)r=2: 12(6−2)r=3: 12(12−6) ⋮r=n: 12(n(n+1)−n(n−1))\begin{align*} r=1: \quad &\, \frac{1}{2}\left(\sqrt{2} - 0\right)\\[4mm] r=2: \quad &\, \frac{1}{2}\left(\sqrt{6} - \sqrt{2}\right)\\[4mm] r=3: \quad &\, \frac{1}{2}\left(\sqrt{12} - \sqrt{6}\right)\\[4mm] &\,\vdots\\[4mm] r=n: \quad &\, \frac{1}{2}\left(\sqrt{n(n+1)} - \sqrt{n(n-1)}\right) \end{align*}

When adding these terms together, all intermediate terms cancel diagonally, leaving only the first negative term and the last positive term:

Sum= 12(n(n+1)−0)= 12n(n+1)\begin{align*} \text{Sum} = &\, \frac{1}{2} \left( \sqrt{n(n+1)} - 0 \right)\\[4mm] = &\, \frac{1}{2} \sqrt{n(n+1)} \end{align*}

(c)

We are given:

∑r=1nkrr(r+1)+r(r−1)= ∑r=1nr\begin{align*} \sum_{r=1}^{n}\frac{kr}{\sqrt{r(r+1)}+\sqrt{r(r-1)}} = &\,\sqrt{\sum_{r=1}^{n}r} \end{align*}

Using the result from (b) on the left-hand side, and the standard sum formula ∑r=1nr=12n(n+1)\sum_{r=1}^{n}r = \frac{1}{2}n(n+1) on the right-hand side:

k(12n(n+1))= 12n(n+1)k2n(n+1)= 12n(n+1)\begin{align*} k \left( \frac{1}{2} \sqrt{n(n+1)} \right) = &\,\sqrt{\frac{1}{2}n(n+1)}\\[4mm] \frac{k}{2} \sqrt{n(n+1)} = &\,\frac{1}{\sqrt{2}} \sqrt{n(n+1)} \end{align*}

Equating the coefficients:

k2= 12k= 22k= 2\begin{align*} \frac{k}{2} = &\,\frac{1}{\sqrt{2}}\\[4mm] k = &\,\frac{2}{\sqrt{2}}\\[4mm] k = &\,\sqrt{2} \end{align*}