(i) $\displaystyle S_{n}=1+2+3+...+n$, then $\displaystyle S_{n}=\frac{n(n+1)}{2}$
⇒ $\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}$
(ii) $\displaystyle S_{n}=1^{2}+2^{2}+3^{2}+...+n^{2}$, then $\displaystyle S_{n}=\frac{n(n+1)(2n+1)}{6}$
⇒ $\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$
(iii) $\displaystyle S_{n}=1^{3}+2^{3}+3^{3}+...+n^{3}$, then $\displaystyle S_{n}=\frac{[n(n+1)]^{2}}{4}$
⇒ $\displaystyle \sum_{k=1}^{n}k^{2}=\frac{[n(n+1)]^{2}}{4}$
⇒ $\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}$
(ii) $\displaystyle S_{n}=1^{2}+2^{2}+3^{2}+...+n^{2}$, then $\displaystyle S_{n}=\frac{n(n+1)(2n+1)}{6}$
⇒ $\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$
(iii) $\displaystyle S_{n}=1^{3}+2^{3}+3^{3}+...+n^{3}$, then $\displaystyle S_{n}=\frac{[n(n+1)]^{2}}{4}$
⇒ $\displaystyle \sum_{k=1}^{n}k^{2}=\frac{[n(n+1)]^{2}}{4}$
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