A lvl H2 Maths: Sequences and Series

TJC 2005 FM P1 Q1

The sequence of real numbers u₁, u₂, u₃, … is such that $U_{r+1}=\small\frac{U_{r}}{1-r^{2}U_{r}}$ .

By using the method of differences, or otherwise, show that $\small\frac{1}{U_{1}}-\frac{1}{U_{N}}=\frac{1}{6}(N-1)(N)(2N-1)$ .

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Answer:

$\small\frac{1}{U_{r+1}}=\frac{1-r^{2}U_{r}}{U_{r}}$
$\small=\frac{1}{U_{r}}-r^{2}$

Hence,
$\small\frac{1}{U_{r}}-\frac{1}{U_{r+1}}=r^{2}$
$\small\sum_{r=1}^{N-1}\left(\frac{1}{U_{r}}-\frac{1}{U_{r+1}}\right)=\sum_{r=1}^{N-1}r^{2}$

$\smallLHS=\sum_{r=1}^{N-1}\left(\frac{1}{U_{r}}-\frac{1}{U_{r+1}}\right)=\left ( \frac{1}{U_{1}}-\frac{1}{U_{2}}\right)+\left ( \frac{1}{U_{2}}-\frac{1}{U_{3}}\right)+...+\left ( \frac{1}{U_{N-1}}-\frac{1}{U_{N}}\right)$
$\small=\frac{1}{U_{1}}-\frac{1}{U_{N}}$

Also,
$\smallRHS=\frac{1}{6}(N-1)N(2(N-1)+1)=\frac{1}{6}(N-1)N(2N-1)$

Hence,
$\small\frac{1}{U_{1}}-\frac{1}{U_{N}}=\frac{1}{6}(N-1)N(2N-1)$ (shown)

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A lvl H2 Maths: Sequences and Series