The first three terms of a sequence are given by $u_1=9$, $u_2=27$ and $u_3=55$. Given that \(

Question

The first three terms of a sequence are given by $u_1=9$, $u_2=27$ and $u_3=55$. Given that $u_n$ is a quadratic polynomial in $n$, find $u_n$ in terms of $n$.[3] It is given $\sum_{r=1}^{n} r^3 = \frac{n^2 (n+1)^2}{4}$ and $u_r = 2r^3 + 5$. Find $\sum_{r=1}^{n} u_r$.[2] Hence, or otherwise, find $\sum_{r=2}^{n} (2(r+2)^3 + 5)$.[3] Explain why the series $\sum_{r=1}^{\infty} u_r$ does not converge.[1]

Accepted Answer

(a) $u_n=5n^2+3n+1$ (b)(i) $\dfrac{n^2(n+1)^2}2+5n$ (b)(ii) $\dfrac{(n+2)^2(n+3)^2}2+5n-77$ (b)(iii) Diverges to $+\infty$.

2025 JPJC P1 Q8

2025 JPJC P1 Q8

Solution:

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