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Question
Find the sum of the following series
62 + 72 + 82 + ... + 212
Solution
62 + 72 + 82 + ... + 212
(12 + 22 + ... + 212) – (12 + 22 + ... + 52)
= `sum_1^21 "n"^2 - sum_1^5 "n"^2`
= `(("n"("n" + 1)(2"n" + 1))/6)_("n" = 21) - (("n"("n" + 1)(2"n" + 1))/6)_("n" = 5)`
= `((21 xx 22 xx 43)/6) - ((5 xx 6 xx 11)/6)`
= 3311 – 55
= 3256
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