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Question
Find the sum of the following series
3 + 6 + 9 + ... + 96
Solution
3 + 6 + 9 + ... + 96
= 3(1 + 2 + 3 + ... + 32)
= `3[sum_1^32 "n"]`
= `3[("n"("n" + 1))/2]_("n" = 32)`
= `3[(32(33))/2]`
= 1584
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