Given that $\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)$, find \(\sum_{r=1}^{n} \left[ \left(

Question

Given that $\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)$, find $\sum_{r=1}^{n} \left[ \left(\frac{1}{2}\right)^r - r\left(r+\frac{1}{3}\right) \right]$, in terms of $n$.[4] Explain why $\sum_{r=1}^{n} \left[ \left(\frac{1}{2}\right)^r - r\left(r+\frac{1}{3}\right) \right] < 1$ for all positive integers $n$.[1]

Accepted Answer

(a) $1 - \left(\frac{1}{2}\right)^n - \frac{n}{3}(n+1)^2$ (b) Since ${n \in \mathbb{Z}^+}$, ${\left(\frac{1}{2}\right)^n > 0}$ and ${\frac{n}{3}(n+1)^2 > 0}$.Thus ${\sum_{r=1}^{n}\left[\left(\frac{1}{2}\right)^r - r\left(r+\frac{1}{3}\right)\right] < 1}$.

2024 HCI Promo Q2

2024 HCI Promo Q2