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IAL 2023 Jan Q2

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 2

(a) Express

1(2n−1)(2n+1)(2n+3)\begin{align*} \frac{1}{(2n - 1)(2n + 1)(2n + 3)}\\[2mm] \end{align*}

in partial fractions.

(2)

(b) Hence, using the method of differences, show that for all integer values of nn ,

∑r=1n1(2r−1)(2r+1)(2r+3)= n(n+2)a(2n+b)(2n+c)\begin{align*} \sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)(2r + 3)} =\,& \frac{n(n + 2)}{a(2n + b)(2n + c)}\\[2mm] \end{align*}

where aa , bb and cc are integers to be determined.

(4)
解答

(a)

Let the expression be written in partial fractions:

1(2n−1)(2n+1)(2n+3)≡ A2n−1+B2n+1+C2n+31≡ A(2n+1)(2n+3)+B(2n−1)(2n+3) +C(2n−1)(2n+1)\begin{align*} \frac{1}{(2n-1)(2n+1)(2n+3)} \equiv &\,\frac{A}{2n-1} + \frac{B}{2n+1} + \frac{C}{2n+3}\\[4mm] 1 \equiv &\,A(2n+1)(2n+3) + B(2n-1)(2n+3)\\[4mm] &\,\hspace{2pt}+ C(2n-1)(2n+1) \end{align*}

Substitute n=12n = \frac{1}{2}:

1= A(2)(4)A= 18\begin{align*} 1 = &\,A(2)(4)\\[4mm] A = &\,\frac{1}{8} \end{align*}

Substitute n=−12n = -\frac{1}{2}:

1= B(−2)(2)B= −14\begin{align*} 1 = &\,B(-2)(2)\\[4mm] B = &\,-\frac{1}{4} \end{align*}

Substitute n=−32n = -\frac{3}{2}:

1= C(−4)(−2)C= 18\begin{align*} 1 = &\,C(-4)(-2)\\[4mm] C = &\,\frac{1}{8} \end{align*}

Therefore, the partial fraction decomposition is:

18(2n−1)−14(2n+1)+18(2n+3)\begin{align*} \frac{1}{8(2n-1)} - \frac{1}{4(2n+1)} + \frac{1}{8(2n+3)} \end{align*}

(b)

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

∑r=1n1(2r−1)(2r+1)(2r+3)= 18∑r=1n(12r−1−22r+1+12r+3)\begin{align*} \sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)(2r+3)} = &\,\frac{1}{8} \sum_{r=1}^{n} \left( \frac{1}{2r-1} - \frac{2}{2r+1} + \frac{1}{2r+3} \right) \end{align*}

Writing out the terms to show the method of differences:

r=1: 11−23+15r=2: 13−25+17r=3: 15−27+19 ⋮r=n−1: 12n−3−22n−1+12n+1r=n: 12n−1−22n+1+12n+3\begin{align*} r=1: \quad &\,\frac{1}{1} - \frac{2}{3} + \frac{1}{5}\\[4mm] r=2: \quad &\,\frac{1}{3} - \frac{2}{5} + \frac{1}{7}\\[4mm] r=3: \quad &\,\frac{1}{5} - \frac{2}{7} + \frac{1}{9}\\[4mm] &\,\vdots\\[4mm] r=n-1: \quad &\,\frac{1}{2n-3} - \frac{2}{2n-1} + \frac{1}{2n+1}\\[4mm] r=n: \quad &\,\frac{1}{2n-1} - \frac{2}{2n+1} + \frac{1}{2n+3} \end{align*}

Summing these terms, all intermediate terms cancel out, leaving:

Sum= 18(1−23+13−12n+1+12n+1−22n+1+12n+3)= 18(1−13−12n+1+12n+3)= 18(23−(2n+3)−(2n+1)(2n+1)(2n+3))= 18(23−2(2n+1)(2n+3))= 18(2(2n+1)(2n+3)−63(2n+1)(2n+3))= 18(2(4n2+8n+3)−63(2n+1)(2n+3))= 18(8n2+16n+6−63(2n+1)(2n+3))= 18(8n(n+2)3(2n+1)(2n+3))= n(n+2)3(2n+1)(2n+3)\begin{align*} \text{Sum} = &\,\frac{1}{8} \left( 1 - \frac{2}{3} + \frac{1}{3} - \frac{1}{2n+1} + \frac{1}{2n+1} - \frac{2}{2n+1} + \frac{1}{2n+3} \right)\\[4mm] = &\,\frac{1}{8} \left( 1 - \frac{1}{3} - \frac{1}{2n+1} + \frac{1}{2n+3} \right)\\[4mm] = &\,\frac{1}{8} \left( \frac{2}{3} - \frac{(2n+3) - (2n+1)}{(2n+1)(2n+3)} \right)\\[4mm] = &\,\frac{1}{8} \left( \frac{2}{3} - \frac{2}{(2n+1)(2n+3)} \right)\\[4mm] = &\,\frac{1}{8} \left( \frac{2(2n+1)(2n+3) - 6}{3(2n+1)(2n+3)} \right)\\[4mm] = &\,\frac{1}{8} \left( \frac{2(4n^2 + 8n + 3) - 6}{3(2n+1)(2n+3)} \right)\\[4mm] = &\,\frac{1}{8} \left( \frac{8n^2 + 16n + 6 - 6}{3(2n+1)(2n+3)} \right)\\[4mm] = &\,\frac{1}{8} \left( \frac{8n(n+2)}{3(2n+1)(2n+3)} \right)\\[4mm] = &\,\frac{n(n+2)}{3(2n+1)(2n+3)} \end{align*}

This is in the required form, with a=3a=3, b=1b=1, and c=3c=3.

解答

(a)

Given the transformation y=1z=z−1y = \frac{1}{z} = z^{-1}, we differentiate with respect to xx:

dydx= −z−2dzdx=−1z2dzdx\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,-z^{-2}\frac{\mathrm{d}z}{\mathrm{d}x} = -\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x} \end{align*}

Substitute yy and dydx\frac{\mathrm{d}y}{\mathrm{d}x} into differential equation (I):

x2(−1z2dzdx)+x(1z)= 2(1z)2−x2z2dzdx+xz= 2z2\begin{align*} x^2\left(-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}\right) + x\left(\frac{1}{z}\right) = &\,2\left(\frac{1}{z}\right)^2\\[4mm] -\frac{x^2}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x} + \frac{x}{z} = &\,\frac{2}{z^2} \end{align*}

Multiply the entire equation by −z2x2-\frac{z^2}{x^2}:

dzdx−zx= −2x2\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} - \frac{z}{x} = &\,-\frac{2}{x^2} \end{align*}

This matches differential equation (II).

(b)

Differential equation (II) is a first-order linear ODE. The integrating factor (IF) is:

IF= e∫−1x dx= e−ln⁡x= 1x\begin{align*} \text{IF} = &\,\mathrm{e}^{\int -\frac{1}{x}\,\mathrm{d}x}\\[4mm] = &\,\mathrm{e}^{-\ln x}\\[4mm] = &\,\frac{1}{x} \end{align*}

Multiply equation (II) by the integrating factor:

1xdzdx−1x2z= −2x3ddx(zx)= −2x3\begin{align*} \frac{1}{x}\frac{\mathrm{d}z}{\mathrm{d}x} - \frac{1}{x^2}z = &\,-\frac{2}{x^3}\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{z}{x}\right) = &\,-\frac{2}{x^3} \end{align*}

Integrate both sides with respect to xx:

zx= ∫−2x−3 dxzx= x−2+czx= 1x2+c\begin{align*} \frac{z}{x} = &\,\int -2x^{-3}\,\mathrm{d}x\\[4mm] \frac{z}{x} = &\,x^{-2} + c\\[4mm] \frac{z}{x} = &\,\frac{1}{x^2} + c \end{align*}

Multiply by xx to find zz:

z= 1x+cx\begin{align*} z = &\,\frac{1}{x} + cx \end{align*}

(c)

Reverse the substitution y=1zy = \frac{1}{z}, so z=1yz = \frac{1}{y}:

1y= 1x+cx\begin{align*} \frac{1}{y} = &\,\frac{1}{x} + cx \end{align*}

Use the condition y=−38y = -\frac{3}{8} when x=3x = 3:

−83= 13+3c−93= 3c−3= 3cc= −1\begin{align*} -\frac{8}{3} = &\,\frac{1}{3} + 3c\\[4mm] -\frac{9}{3} = &\,3c\\[4mm] -3 = &\,3c\\[4mm] c = &\,-1 \end{align*}

Substitute c=−1c = -1 back into the equation:

1y= 1x−x1y= 1−x2xy= x1−x2\begin{align*} \frac{1}{y} = &\,\frac{1}{x} - x\\[4mm] \frac{1}{y} = &\,\frac{1 - x^2}{x}\\[4mm] y = &\,\frac{x}{1 - x^2} \end{align*}