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Question
Find \[\displaystyle\sum_{r=1}^{n}\frac{1 + 2 + 3 + ... + r}{r}\]
Solution
\[\displaystyle\sum_{r=1}^{n}\frac{1 + 2 + 3 + ...+ r}{r}\]
= \[\displaystyle\sum_{r=1}^{n}\frac{r(r + 1)}{2r}\]
= \[\frac{1}{2}\displaystyle\sum_{r=1}^{n}(r + 1)\]
`= 1/2 [sum_(r=1)^n "r" + sum_(r=1)^n 1]`
= `1/2[("n"("n" + 1))/2 + "n"]`
= `"n"/4[("n" + 1) + 2]`
= `"n"/4("n" + 3)`
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