සිතන්නට යමක්…..
\frac x1+\frac x{1+2}+\frac x{1+2+3}+…+\frac x{1+2+…+4041}=4041\;\text{විසඳන්න.}
\begin{array}{rcl}{\mathrm U}_{\mathrm r}\;&=&\;1\;+\;2\;+\;3\;+.......+\;\mathrm r\;\text{ ශ්රේණියේ පද r හි ඓක්ය}\\&=&\frac{\mathrm r\;(1+\;\mathrm r\;)}2\end{array}
\begin{array}{l}\text{(පොදු අන්තරය 1 වූද පළමු පදය 1 වූද සමාන්තර ශ්රේණියක පද r හි ඓක්ය Sr = පද ගණන(ආරම්භක පදය+අවසාන පදය)/2 මඟින් ලැබේ.)}\\\\\mathrm{පොදු}\;\mathrm{පදය}\;\;=\;\frac{\mathrm x}{{\mathrm U}_{\mathrm r}}\;=\;\frac{2\mathrm x}{\mathrm r(\mathrm r\;+\;1)}\;=\;2\mathrm x\frac{(\mathrm r\;+\;1)-\;\mathrm r}{\mathrm r(\mathrm r\;+\;1)}\;=\;\frac{2\mathrm x}{\mathrm r}\;=\;\frac{2\mathrm x}{\mathrm r\;+\;1}\;=\;\mathrm f\;(\mathrm r)\;-\;\mathrm f(\mathrm r\;+\;1)\end{array}
\begin{array}{rcl}\overset{4041}{\underset{\mathrm r\;=\;1}{\sum\;\;}}\;\frac{\displaystyle\mathrm x}{\displaystyle{\mathrm U}_{\mathrm r}}\;&=&\;\overset{4041}{\underset{\mathrm r\;=\;1}{\sum\;\;}}\;\mathrm f(\mathrm r)\;-\;\mathrm f(\mathrm r+1)\\&=&\;\mathrm f(1)\;-\;\mathrm f(2)\\&=&\;\mathrm f(2)\;-\;\mathrm f(3)\\&&.\\&&.\\&&.\\&=&\;\mathrm f(4041)\;-\;\mathrm f(4042)\\&&.................................\\&&\\\overset{4041}{\underset{\mathrm r\;=\;1}{\sum\;\;}}\;\frac{\mathrm x}{{\mathrm U}_{\mathrm r}}\;&=&\;\mathrm f(1)\;-\;\mathrm f(2)\\[4px]4041\;\;&=&\;2\mathrm x\;-\;\frac{2\mathrm x}{4042}\\[4px]4041\;&=&\;2\mathrm x\;\left(\frac{4042}{4042}\;-\;\frac1{4042}\right)\\[4px]4041\;&=&\;2\mathrm x\;\left(\frac{\displaystyle4041}{\displaystyle4042}\right)\\2x\;&=&\;4042\\x\;&=&\;2021\end{array}
