Given that ${{u}_{r}}=\frac{1}{{{r}^{3}}}$, Show that ${{u}_{r-1}}-{{u}_{r+1}}=\frac{6{{r}^{2}}+2}{{

Question

Given that ${{u}_{r}}=\frac{1}{{{r}^{3}}}$, Show that ${{u}_{r-1}}-{{u}_{r+1}}=\frac{6{{r}^{2}}+2}{{{\left( {{r}^{2}}-1 \right)}^{3}}}$. Hence find $\sum\limits_{r=2}^{n}{\frac{3{{r}^{2}}+1}{{{\left( {{r}^{2}}-1 \right)}^{3}}}}$ in terms of $n$. Use your answer to part (ii) to find $\sum\limits_{r=3}^{n}{\frac{3{{\left( r-1 \right)}^{2}}+1}{{{\left[ {{\left( r-1 \right)}^{2}}-1 \right]}^{3}}}}$ in terms of $n$. Hence find $\sum\limits_{r=3}^{\infty }{\frac{3{{\left( r-1 \right)}^{2}}+1}{{{\left[ {{\left( r-1 \right)}^{2}}-1 \right]}^{3}}}}$.

Accepted Answer

(ii) $\frac12\left(\frac98-\frac1{n^3}-\frac1{(n+1)^3}\right)$; (iii) $\frac9{16}$

2009 SAJC P2 Q3 [Modified]

2009 SAJC P2 Q3 [Modified]

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