Summation and series: DHS 2024 Promo Q9

(a)(i) $$\begin{aligned} \sum_{r=0}^n\frac{(x+3)^r}{4^{r+1}} &=\frac14\sum_{r=0}^n\left(\frac{x+3}{4}\right)^r\\ &=\frac14\sum_{r=0}^n\left[\left(\frac{x+3}{4}\right)^0+\left(\frac{x+3}{4}\right)^1+\cdots+\left(\frac{x+3}{4}\right)^n\right]\\ &=\frac14\left[\frac{1-\left(\frac{x+3}{4}\right)^{n+1}}{1-\frac{x+3}{4}}\right]\\ &=\frac14\left[\frac{1-\left(\frac{x+3}{4}\right)^{n+1}}{\frac{1-x}{4}}\right]\\ &=\frac1{1-x}\left[1-\left(\frac{x+3}{4}\right)^{n+1}\right] \end{aligned}$$ (a)(ii) Common ratio, $r$ of G.P. $=\dfrac{x+3}{4}$

When $x=-5$, $r=\dfrac{-5+3}{4}=-\dfrac12$.

Since $|r|=\dfrac12<1$, the G.P. converges. Hence, the series $\displaystyle\sum_{r=0}^n\dfrac{(x+3)^r}{4^{r+1}}$ converges. $$\begin{aligned} \lim_{n\to\infty}\sum_{r=0}^n\frac{(-5+3)^r}{4^{r+1}} &=\lim_{n\to\infty}\frac1{1-(-5)}\left[1-\left(\frac{-5+3}{4}\right)^{n+1}\right]\\ &=\lim_{n\to\infty}\frac16\left[1-\left(-\frac12\right)^{n+1}\right]\\ &=\frac16 \end{aligned}$$ (b)(i) $$\begin{aligned} \sum_{r=6}^{2k}r(3r-2) &=\sum_{r=6}^{2k}(3r^2-2r)\\ &=3\left[\sum_{r=1}^{2k}r^2-\sum_{r=1}^{5}r^2\right]-2\sum_{r=6}^{2k}r\\ &=3\left[\frac{2k}{6}(2k+1)(4k+1)-\frac56(6)(11)\right] -2\left(\frac{2k-6+1}{2}(6+2k)\right)\\ &=k(2k+1)(4k+1)-165-2(2k-5)(3+k)\\ &=k(2k+1)(4k+1)-2(2k-5)(k+3)-165 \end{aligned}$$