The $n$ th term of a series $G$ is given by $g_n = 3n^2 - 17n + 17$, where $n \geq 1$.

Question

The $n$ th term of a series $G$ is given by $g_n = 3n^2 - 17n + 17$, where $n \geq 1$. Given that $\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)$, find $\sum_{r=m+1}^{n} g_r$. (You need not simplify your answer.)[4] The $n$th term of a series $H$ is given by $h_n = 7(5^{n-3}) + 10$, where $n \geq 1$. State the smallest number that can be found in both series $H$ and $G$.[1] The sum of the first $n$ terms of a series $J$ is given by $\frac{b}{2}(3^n - 1)$, where $b$ is a positive odd integer. By finding the $n$th term of series $J$, explain why each of the terms in series $J$ is a term of the arithmetic progression with first term $1$ and common difference $2$.[3]

Accepted Answer

(a) $\dfrac n2(n+1)(2n+1)-\dfrac m2(m+1)(2m+1)+\dfrac{17}{2}(n-m)(1-m-n)$ (b) $45$ (c) Explained

2025 NYJC P1 Q5

2025 NYJC P1 Q5

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