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) $1-\left(\dfrac12\right)^n-\dfrac{n(n+1)^2}{3}$ (b) The sum is less than $1$ for every positive integer $n$

2024 HCI Promo Q2

2024 HCI Promo Q2

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