Summation and series: HCI 2024 Promo Q2

(a)

Given that $\displaystyle\sum_{r=1}^{n}r^2=\frac n6(n+1)(2n+1)$, find $\displaystyle\sum_{r=1}^{n}\left[\left(\frac12\right)^r-r\left(r+\frac13\right)\right]$, in terms of $n$. [4]

Suggested Solutions

(a)

$$\begin{aligned} \sum_{r=1}^{n}\left[\left(\frac{1}{2}\right)^{\!r} - r\left(r + \frac{1}{3}\right)\right] &= \sum_{r=1}^{n}\left(\frac{1}{2}\right)^{\!r} - \sum_{r=1}^{n} r^2 - \frac{1}{3}\sum_{r=1}^{n} r\\ &= \frac{\frac{1}{2}\left[1 - \left(\frac{1}{2}\right)^{\!n}\right]}{1 - \frac{1}{2}} - \frac{n}{6}(n+1)(2n+1) - \frac{n}{6}(1 + n)\\ &= 1 - \left(\frac{1}{2}\right)^{\!n} - \frac{n}{6}(n+1)(2n + 1 + 1)\\ &= 1 - \left(\frac{1}{2}\right)^{\!n} - \frac{n}{3}(n+1)^2. \end{aligned}$$

(b)

Explain why $\displaystyle\sum_{r=1}^{n}\left[\left(\frac12\right)^r-r\left(r+\frac13\right)\right]<1$ for all positive integers $n$. [1]

Suggested Solutions

(b)

Since $n\in\mathbb Z^+$, $\left(\dfrac12\right)^n>0$ and $\dfrac n3(n+1)^2>0$. $$\therefore\ \sum_{r=1}^{n}\left[\left(\frac12\right)^r-r\left(r+\frac13\right)\right]<1.$$