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प्रश्न
Find `sum_("r" = 1)^"n"((1 + 2 + 3 .... + "r")/"r")`
उत्तर
`sum_("r" = 1)^"n"((1 + 2 + 3 .... + "r")/"r")`
= `sum_("r" = 1)^"n" ("r"("r" + 1))/(2"r")`
= `1/2 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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