Given that $\sum_{r=1}^{n} r^{2} = \frac{n}{6}(n+1)(2n+1)$, evaluate \(\sum_{r=-n}^{n} (r+1)(

Question

Given that $\sum_{r=1}^{n} r^{2} = \frac{n}{6}(n+1)(2n+1)$, evaluate $\sum_{r=-n}^{n} (r+1)(r+3)$ in terms of $n$.[4] Using standard series from the List of Formulae (MF27), find the range of values of $x$ for which the series $\sum_{r=1}^{\infty} \frac{(-1)^{r+1} x^{r}}{r 2^{r}}$ converges. State the sum to infinity in terms of $x$.[3]

Accepted Answer

(a) $\dfrac{(2n+1)(n^2+n+9)}3$ (b) $-2<x\le2$, $S_\infty=\ln\left(1+\dfrac x2\right)$

2024 DHS Promo Q3

2024 DHS Promo Q3

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