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प्रश्न
Find \[\displaystyle\sum_{r = 1}^{n}\frac{1 + 2 + 3 + ... + r}{r}\]
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उत्तर
\[\displaystyle\sum_{r=1}^{n}\frac{1 + 2 + 3 + ...+ r}{r}\]
= \[\displaystyle\sum_{r=1}^{n}\frac{r(r + 1)}{2r}\]
= \[\frac{1}{2}\displaystyle\sum_{r=1}^{n}(r + 1)\]
`= 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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