For this question, you may use these results: \[\sum_{r=1}^{n}r^2 = \frac{n(n+1)(2n+1)}{6} \quad \te

Question

For this question, you may use these results: $$\sum_{r=1}^{n}r^2 = \frac{n(n+1)(2n+1)}{6} \quad \text{and} \quad \sum_{r=1}^{n}r^3 = \frac{n^2 (n+1)^2}{4}.$$ Show that $\sum_{r=1}^{n}r(r+1)^2 = \frac{n(n+1)(n+2)(3n+5)}{12}$.[3] Hence find $\sum_{r=5}^{n-1}(r+2)(r+3)^2$ in terms of $n$.[3] The sequence $u_1, u_2, u_3, \dots$ 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

(a)(ii) $\frac{(n+1)(n+2)(n+3)(3n+8)}{12}-644$. (b) $u_2=-1$, $u_3=\frac12$, $u_4=2$, and $u_{2025}=\frac12$.

2025 ACJC P2 Q3

2025 ACJC P2 Q3

Video Solution:

Solution:

Solution locked