For this question, you may use these results: $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{n(n+1)(2n+1)}{

Question

For this question, you may use these results: $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{n(n+1)(2n+1)}{6}$ and $\sum\limits_{r=1}^{n}{{{r}^{3}}}=\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}$. Show that $\sum\limits_{r=1}^{n}{r}{{(r+1)}^{2}}=\frac{n(n+1)(n+2)(3n+5)}{12}$. [3] Hence find $\sum\limits_{r=5}^{n-1}{(r+2)}{{(r+3)}^{2}}$ in terms of $n$. [3] The sequence ${{u}_{1}},{{u}_{2}},{{u}_{3}},\ldots $ is defined by ${{u}_{1}}=2$, ${{u}_{n+1}}=\frac{1}{1-{{u}_{n}}}$, $n\ge 1$. Find the values of ${{u}_{2}}$, ${{u}_{3}}$ and ${{u}_{4}}$. Hence find the value of ${{u}_{2025}}$. [3]

Accepted Answer

$\frac{n(n+1)(n+2)(3n+5)}{12}$, $\frac{(n+1)(n+2)(n+3)(3n+8)}{12} - 644$, $u_{2025} = \frac{1}{2}$

2025 ACJC P2 Q3

2025 ACJC P2 Q3

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