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Question
Find the sum of the following series
103 + 113 + 123 + ... + 203
Solution
103 + 113 + 123 + ... + 203
= (13 + 23+ 33 + ... + 203) – (13 + 23 + 33 + ... + 93)
= `sum_1^20 "n"^3 - sum_1^9 "n"^3`
= `(("n"("n" + 1))/2)_("n" = 20)^2 - (("n"("n" + 1))/2)_("n" = 9)^2`
= `((20 xx 21)/2)^2 - ((9 xx 10)/2)^2`
= 2102 – 452
= 44100 – 2025
= 42075
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