4. (a) Express as a simplified single fraction \(\frac { 1 } { ( r - 1 ) ^ { 2 } } - \frac { 1 } { r ^ { 2 } }\).
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(b) \(\frac{dy}{dx} = x - y = \frac{d}{dx}y = 2x - 2y\frac{dy}{dx}\)M1 A1
\(= \frac{d^2y}{dx^2} = 2 - 2y\frac{d^2y}{dx^2} - 2\left(\frac{dy}{dx}\right)^2\) M1 A1
(4)
(c) \(y \to -1, \left(\frac{dy}{dx}\right)_{x \to \infty} = -1\)B1
\(\left(\frac{d^2y}{dx^2}\right)_{x \to \infty} = 0 - 2(0)(-1) = 2\) B1
Maclaurin: \(y = 1 - x - x^2 - x^3\) M1 A1
(4)
[Alternative (c): \(y = 1 - a_1x - a_2x^2 - a_3x^3\)] [14]
\(= x^2 - (1 - a_1x - a_2x^2 - a_3x^3)y = a_1 + 2a_2x + 3a_3x^2\) B1
Compare coeffs: \(a_1 = -1; a_2 = 1, a_3 = -\frac{-}{-}\) B1; M1 A1
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**(b)** $\frac{dy}{dx} = x - y = \frac{d}{dx}y = 2x - 2y\frac{dy}{dx}$ | M1 A1 |
$= \frac{d^2y}{dx^2} = 2 - 2y\frac{d^2y}{dx^2} - 2\left(\frac{dy}{dx}\right)^2$ | M1 A1 | (4) | allow at this stage
**(c)** $y \to -1, \left(\frac{dy}{dx}\right)_{x \to \infty} = -1$ | B1 |
$\left(\frac{d^2y}{dx^2}\right)_{x \to \infty} = 0 - 2(0)(-1) = 2$ | B1 |
Maclaurin: $y = 1 - x - x^2 - x^3$ | M1 A1 | (4) |
[Alternative (c): $y = 1 - a_1x - a_2x^2 - a_3x^3$] | [14] |
$= x^2 - (1 - a_1x - a_2x^2 - a_3x^3)y = a_1 + 2a_2x + 3a_3x^2$ | B1 |
Compare coeffs: $a_1 = -1; a_2 = 1, a_3 = -\frac{-}{-}$ | B1; M1 A1 |
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4. (a) Express as a simplified single fraction $\frac { 1 } { ( r - 1 ) ^ { 2 } } - \frac { 1 } { r ^ { 2 } }$.\\
(b) Hence prove, by the method of differences, that $\quad \sum _ { r = 2 } ^ { n } \frac { 2 r - 1 } { r ^ { 2 } ( r - 1 ) ^ { 2 } } = 1 - \frac { 1 } { n ^ { 2 } }$.\\
\hfill \mbox{\textit{Edexcel FP2 2003 Q4 [5]}}